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MATHEMATICAL   TEXT-BOOKS 

BY 

GEORGE  A.  WENTWORTH 


Elementary  Arithmetic 

Practical  Arithmetic 

Mental  Arithmetic 

Primary  Arithmetic    (Wentworth  and  Reed) 

Grammar  School  Arithmetic 

Advanced  Arithmetic 

Exercises  in  Arithmetic    (Wentworth  and  Hill) 

First  Steps  in  Algebra 

New  School  Algebra 

Elementary  Algebra 

Elements  of  Algebra 

Shorter  Course  in  Algebra 

College  Algebra  (Revised  Edition) 

Advanced  Algebra 

Complete  Algebra 

Higher  Algebra 

Exercises  in  Algebra     (Wentworth  and  Hill) 

First  Steps  in  Geometry     (Wentworth  and  Hill) 

Plane  and  Solid  Geometry  (Revised) 

Plane  Geometry  (Revised) 

Solid  Geometry  (Revised) 

Plane  and  Solid  Geometry  and  Plane  Trigonometry 

(Second  Revised  Edition) 
Analytic  Geometry 
Logarithms,  Metric  Measures,  etc. 
Geometrical  Exercises 
Syllabus  of  Geometry 

Examination  Manual  in  Geometry   (Wentworth  and  Hill) 
Exercise  Manual  in  Geometry    (Wentworth  and  Hill) 
Plane  Trigonometry  (Second  Revised  Edition) 
Plane   Trigonometry    and    Tables    (Second   Revised 

Edition) 
Plane  and  Spherical  Trigonometry  (Second  Revised 

Edition) 
Plane  and  Spherical   Trigonometry  and  Tables 

(Second  Revised  Edition) 
Plane  Trigonometry,  Surveying,  and  Tables  (Second 

Revised  Edition) 
Surveying  and  Tables  (Second  Revised  Edition) 
Plane  and  Spherical  Trigonometry,  Surveying,  and 

Tables  (Second  Revised  Edition) 
Plane  and  Spherical  Trigonometry,  Surveying,  and 

Navigation  (Second  Revised  Edition) 
Logarithmic  and  Trigonometric  Tables 
Seven  Tables,  Complete  (Wentworth  and  mil) 


NEW 


'    SCHOOL   ALQEBBA 


BY 


G.  A.  WENTWOKTH 

AlTTHOB  OF  A  SERIES  OF  TEXT-BOOKS  IN  MATHEMATICS 


GINN  AND  COMPANY 

BOSTON     •     NEW   YORK     •     CHICAGO     •     LONDON 
ATLANTA     •     DALLAS     •     COLUMBUS     •     SAN    FRANCISCO 


Copyright,  1898,  by 
GEORGE  A.  WENTWORTH 


ALL  BIGHTS  RESERVED 
A  822.5 


TEfte   satftenatum   jgrcgg 

GINN  AND  COMPANY-  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PEEFAOE. 


The  first  chapter  of  this  book  prepares  the  way  for  quite 
a  full  treatment  of  simple  integral  equations  with  one 
unknown  number.  In  the  first  two  chapters  only  positive 
numbers  are  involved,  and  the  beginner  is  led  to  see  the 
practical  advantages  of  Algebra  before  he  encounters  the 
difficulties  of  negative  numbers. 

The  definitions  and  explanations  contained  in  these 
chapters  should  be  carefully  read  at  first ;  after  the  learner 
has  become  familiar  with  algebraic  operations,  special  atten- 
tion should  be  given  to  the  principal  definitions. 

The  third  chapter  contains  a  simple  explanation  of  nega- 
tive numbers.  The  recognition  of  the  fact  that  the  real 
nature  of  subtraction  is  counting  backwards,  and  that  the 
real  nature  of  multiplication  is  forming  the  product  from 
the  multiplicand  precisely  as  the  multiplier  is  formed  from 
unity,  makes  an  easy  road  to  the  laws  of  addition  and  sub- 
traction of  algebraic  numbers,  and  to  the  law  of  signs  in 
multiplication  and  division.  All  the  principles  and  rules 
of  this  chapter  are  illustrated  and  enforced  by  numerous 
examples  involving  simple  algebraic  expressions  only. 

The  ordinary  processes  with  compound  expressions,  in- 
cluding cases  of  resolution  into  factors,  and  the  treatment 
of  fractions,  naturally  follow  the  third  chapter.  The  im- 
mediate succession  of  topics  that  require  similar  work  is  of 
the  highest  importance  to  the  beginner,  and  it  is  hoped  that 
the  chapters  on  compound  expressions  will  prove  interest- 
ing, and  give  sufficient  readiness  in  the  use  of  symbols. 


53905 


iv  PREFACE. 

The  chapter  on  Factors  has  been  made  as  complete  as 
possible  for  an  elementary  text-book,  with  a  view  to  shorten 
subsequent  work.  The  easy  method  of  resolving  quadratic 
trinomials  into  factors,  whether  the  coefficient  of  the  square 
of  the  letter  involved  is  unity  or  greater  than  unity,  and  the 
Factor  Theorem,  explained  on  page  102,  will  be  found  of  very 
great  service  in  abridging  algebraic  processes.  Examples 
of  short  methods  for  finding  the  highest  common  factor  of 
compound  expressions  are  given  on  page  118 ;  and  examples 
of  short  methods  for  solving  quadratic  equations  by  resolv- 
ing them  into  factors  are  given  on  pages  272  and  273. 

A  five-place  table  of  logarithms  is  placed  at  the  end  of 
the  book  instead  of  a  four-place  table.  Five-place  loga- 
rithms are  in  common  use  for  practical  calculations,  and 
are  required  by  most  colleges  and  science  schools  for  the 
solution  of  problems  set  in  entrance  examination  papers. 

The  exercises  throughout  the  book  are  carefully  graded. 
They  are  sufficiently  varied  and  interesting,  and  are  not  so 
difficult  as  to  discourage  the  learner,  or  so  easy  as  to  deprive 
him  of  the  satisfaction  of  well-earned  success. 

The  author  has  spared  no  pains  to  make  this  a  model 
text-book  in  subject-matter  and  mechanical  execution.  The 
remarkable  favor  with  which  his  other  Algebras  have  been 
received  is  shown  by  the  fact  that  nearly  a  million  copies 
have  already  been  sold,  and  the  sale  continues  to  increase 
from  year  to  year.  The  author  trusts  that  this  new  candi- 
date for  favor  will  have  the  same  generous  reception,  and  be 
found  to  meet  fully  the  requirements  of  the  recent  advance 
in  the  science  and  method  of  teaching  Elementary  Algebra. 

The  author  is  under  obligations  to  many  teachers  for 

valuable  suggestions,  and  he  will  be  thankful  for  corrections 

or  criticisms. 

G.  A.  WENTWOETH. 

Exeter,  N.  H.,  June,  1898. 


CONTENTS. 


CHAPTER  PAGE 

I.    Definitions  and  Notation 1 

II.    Simple  Equations 15 

III.  Positive  and  Negative  Numbers       ....  33 

IV.  Addition  and  Subtraction 49 

V.    Multiplication  and  Division     .         .         .        .         .69 

VI.    Special  Rules 76 

VII.    Factors 85 

VIII.    Common  Factors  and  Multiples    ....  108 

IX.    Fractions 123 

X.    Fractional  Equations 148 

XL    Simultaneous  Simple  Equations        .         .         .         .174 

XII.    Problems  with   Two  or  More    Unknown  Numbers  190 

XIII.  Simple  Indeterminate  Equations      ....  205 

XIV.  Inequalities 208 

XV.   Involution  and  Evolution          *.     .   .  212 

XVI.    Theory  of  Exponents     ...........  230 

XVII.    Radical  Expressions 238 

XVIII.    Imaginary  Expressions .    ,    256 

XIX.    Quadratic  Equations  .     ,   .         .     .   .     ...     .    .     .    .  262 

XX.    Simultaneous  Quadratics 293 

XXI.    Ratio,  Proportion,  and  Variation  ....  303 

XXII.    Progressions 322 

XXIII.  Variables  and  Limits 338 

XXIV.  Properties  of  Series 345 

XXV.    Binomial  Theorem 352 

XXVI.    Logarithms 372 

XXVII.    Graphs 409 


NOTICE    TO    TEACHERS. 


Pamphlets  containing  the  answers  will  be  furnished 
without  charge  to  teachers  for  their  classes,  on  appli- 
cation to  G-INN  £  COMPANY,  Publishers. 


NEW    SCHOOL    ALGEBEA. 

CHAPTER  I. 
DEFINITIONS  AND  NOTATION. 


Numbers  and  Number-Symbols. 

1.  Algebra.     Algebra,  like  Arithmetic,  treats  of  numbers. 

2.  Units.  In  counting  separate  objects  or  in  measuring 
magnitudes,  the  standards  by  which  we  count  or  measure 
are  called  units. 

Thus,  in  counting  the  boys  in  a  school,  the  unit  is  a  boy ;  in  selling 
eggs  by  the  dozen,  the  unit  is  a  dozen  eggs ;  in  selling  bricks  by  the 
thousand,  the  unit  is  a  thousand  bricks ;  in  expressing  the  measure  of 
short  distances,  the  unit  is  an  inch,  a  foot,  or  a  yard ;  in  expressing 
the  measure  of  long  distances,  the  unit  is  a  rod,  or  a  mile. 

3.  Numbers.  Repetitions  of  the  unit  are  expressed  by 
numbers. 

A  single  unit  and  groups  of  units  formed  by  successive  additions  of 
a  unit  may  be  represented  as  follows : 

/    //    ///    ////    mi   *  mi  i 
mil    mi in    iw mi    iw mi 

These  representative  groups  are  named  one,  two,  three,  four,  five, 
six,  seven,  eight,  nine,  ten;  and  are  known  collectively  under  the 
general  name  of  numbers.  It  is  obvious  that  these  representative 
groups  will  have  the  same  meaning,  whatever  the  units  may  be  that 
are  counted. 


2  DEFINITIONS  AND  NOTATION. 


4.  Quantities.  A  number  of  specified  units  of  any  kind 
is  called  a  quantity  ;   as  4  pounds,  5  oranges. 

Note.  Quantities  are  often  called  concrete  numbers,  the  adjective 
concrete  being  transferred  from  the  units  counted  to  the  numbers 
that  count  them;  but  a  number  signifies  the  times  a  unit  is  taken, 
whether  the  unit  is  expressed  or  understood,  and  is  always  abstract. 

Thus,  4  barrels  of  flour  means  4  times  1  barrel  of  flour ;  and  10 
cords  of  wood  means  10  times  1  cord  of  wood. 

5.  Number-Symbols  in  Arithmetic.  Instead  of  groups  of 
straight  marks,  we  use  in  Arithmetic  the  arbitrary  symbols 
1,  2,  3,  4,  5,  6,  7,  8,  9,  called  Arabic  numerals,  for  the  num- 
bers one,  two,  three,  four,  five,  six,  seven,  eight,  nine. 

The  next  number,  ten,  is  indicated  by  writing  the  figure 
1  in  a  different  position,  so  that  it  shall  signify  not  one,  but 
ten.  This  change  of  position  is  effected  by  introducing  a 
new  symbol,  0;  called  nought  or  zero,  and  signifying  none. 

All  succeeding  numbers  up  to  the  number  consisting  of  10  tens  are 
expressed  by  writing  the  figure  for  the  number  of  tens  they  contain 
in  the  second  place  from  the  right,  and  the  figure  for  the  number  oi 
units  besides  in  the  first  place.  The  hundreds  of  a  number  are  written 
in  the  third  place  from  the  right.  The  thousands  are  written  in  the 
fourth  place  from  the  right ;  and  so  on. 

6.  Number-Symbols  in  Algebra.  Algebra  employs  the 
letters  of  the  alphabet  in  addition  to  the  figures  of  Arith- 
metic to  represent  numbers.  The  letters  of  the  alphabet 
are  used  as  general  symbols  of  numbers  to  which  any  par- 
ticular values  may  be  assigned.  In  any  problem,  however, 
a  letter  is  understood  to  have  the  same  value  throughout 
the  problem. 

7.  Terms  Common  to  Arithmetic  and  Algebra.  Terms  com- 
mon to  Arithmetic  and  Algebra,  as  addition,  sum,  subtrac- 
tion, minuend,  subtrahend,  difference,  etc.,  have  the  same 
meaning  in  both ;  or  an  extended  meaning  in  Algebra  con- 
sistent with  the  sense  attached  to  them  in  Arithmetic. 


DEFINITIONS  AND  NOTATION.  3 

The  Principal  Signs  of  Operations. 

The  principal  signs  of  operations  are  the  same  in  Algebra 
as  in  Arithmetic. 

8.  The  Sign  of  Addition,  +.     The  sign  +  is  read  plus. 

Thus,  4  +  3  is  read  4  plus  3,  and  indicates  that  the  number  3  is  to 
be  added  to  the  number  4  ;  a  -f-  b  is  read  a  plus  6,  and  indicates  that 
the  number  6  is  to  be  added  to  the  number  a. 

9.  The  Sign  of  Subtraction,  — .    The  sign  —  is  read  minus. 

Thus,  4  —  3  is  read  4  minus  3,  and  indicates  that  the  number  3  is 
to  be  subtracted  from  the  number  4 ;  a  —  b  is  read  a  minus  6,  and 
indicates  that  the  number  b  is  to  be  subtracted  from  the  number  a. 

10.  The  Sign  of  Multiplication,  X.  The  sign  X  is  read 
times,  or  multiplied  by. 

Thus,  4  X  8  is  read  4  times  3,  and  indicates  that  the  number  3  is 
to  be  multiplied  by  4  ;  a  X  6  is  read  a  times  b,  and  indicates  that  the 
number  6  is  to  be  multiplied  by  the  number  a. 

A  dot  is  sometimes  used  for  the  sign  of  multiplication. 
Thus,  2  •  3  •  4  •  5  means  the  same  as  2  X  3  X  4  X  5.  Either 
sign  is  read  multiplied  by  when  followed  by  the  multiplier. 
$a  X  b,  or  $a  •  b,  is  read  a  dollars  multiplied  by  b. 

11.  The  Sign  of  Division,  -J-.    The  sign  ■+-  is  read  divided  by. 

Thus,  4  -r  2  is  read  4  divided  by  2,  and  indicates  that  the  number 
4  is  to  be  divided  by  2 ;  a  -r  6  is  read  a  divided  by  6,  and  indicates 
that  the  number  a  is  to  be  divided  by  the  number  b. 

Division  is  also  indicated  by  writing  the  dividend  above 
the  divisor  with  a  horizontal  line  between  them;  or  by 
separating  the  dividend  from  the  divisor  by  an  oblique 
line,  called  the  solidus. 

Thus,  -i  or  a fby  means  the  same  as  a  •+-  b. 

Note.  The  operation  of  adding  b  to  a,  of  subtracting  6  from  a,  of 
multiplying  a  by  6,  or  of  dividing  a  by  6  is  algebraically  complete 
when  the  two  letters  are  connected  by  the  proper  sign. 


4  DEFINITIONS  AND  NOTATION 

12.  The  Radical  Sign,  ->/•  Tne  sign  V  is  called  the 
radical  sign,  and  denotes  that  a  root  of  the  number  before 
which  it  is  placed  is  to  be  found. 

Other  Signs  Used  in  Algebra. 

13.  The  Sign  of  Equality,  =.  The  sign  =  is  read  is 
equal  to,  and  when  placed  between  two  numbers  indicates 
that  these  two  numbers  are  equal. 

Thus,  8  +  4  =  12  means  that  the  sum  of  8  and  4  is  equal  to  12 ; 
x  +  y  =  20  means  that  the  sum  of  x  and  y  is  equal  to  20 ;  and  x  =  a  +  b 
means  that  x  is  equal  to  the  sum  of  a  and  b. 

14.  The  Sign  of  Deduction,  .•.  The  sign  .'.is  read  hence 
or  therefore. 

15.  The  Sign  of  Continuation,  The  sign is  read 

and  so  on.  * 

Thus,  1,  2,  3,  4, is  read  one,  two,  three,  four,  and  so  on.     oi, 

Oa,  as,  an  is  read,  a  sub  one,  a  sub  two,  a  sub  three,  and  so  on  to 

a  sub  n.     a',  a",  a'", is  read  a  prime,  a  second,  a  third,  and  so  on. 

16.  The  Signs  of  Aggregation.  The  signs  of  aggregation 
are  the  parenthesis  (  ),  the  bracket  [  ],  the  brace  \  \,  the 
vinculum  — ,  and  the  bar  | . 

These  signs  mean  that  the  indicated  operations  in  the 
expressions  affected  by  them  are  to  be  performed  first,  and 
the  result  treated  as  a  single  number. 

Thus,  (a  +  b)  X  c  means  that  the  sum  of  a  and  b  is  to  be  multiplied 
by  c ;  (a  —  b)  X  c  means  that  the  difference  of  a  and  b  is  to  be  multi- 
plied by  c. 

The  vinculum  is  written  over  the  expression  that  is  to  be 
treated  as  a  single  number. 

Thus,  a  —  b-rc  means  the  same  as  a  —  (b  +  c),  and  signifies  that 
the  sum  of  b  and  c  is  to  be  subtracted  from  a ;  and  Va  —  b  means 
the  same  as  V(#  ~~  &)>  and  signifies  that  b  is  to  be  subtracted  from  a, 
and  the  square  root  of  the  remainder  found. 


DEFINITIONS  AND  NOTATION.  5 

Factors,  Powers,  Roots. 

17.  Factors.  When  a  number  is  the  product  of  two  01 
more  numbers,  each  of  these  numbers,  or  the  product  of 
two  or  more  of  them,  is  called  a  factor  of  the  given  number. 

Thus,  2,  a,  b,  2  a,  2  6,  ab  are  factors  of  2  ab. 

18.  Factors  expressed  by  letters  are  called  literal  factors  j 
factors  expressed  by  figures  are  called  numerical  factors. 

19.  The  sign  X  is  omitted  between  factors,  if  the  factors 
are  letters,  or  a  numerical  factor  and  a  literal  factor. 

Thus,  we  write  63  ab  for  63  x  a  x  b ;  we  write  abc  for  a  X  b  X  c. 

20.  The  expression  abc  must  not  be  confounded  with 
a  +  b  -f-  c.     afo  is  a  product ;   a  +  6  +-  c  is  a  sum. 

If  a  =  2,  b  =  3,  c  =  4, 

then  afo  =  2  X  3  X  4  =  24 ; 

but  a  +  &  +  c  =  2  +  3  +  4  =  9. 

Note.  When  a  sign  of  operation  is  omitted  in  the  potation  of 
Arithmetic,  it  is  always  the  sign  of  addition;  but  when  a  sign  of 
operation  is  omitted  in  the  notation  of  Algebra,  it  is  always  the  sign 
of  multiplication.  Thus,  456  means  400  +  50  +  6,  but  4  ab  means 
4  X  a  X  6. 

21.  If  one  factor  of  a  product  is  equal  to  0,  the  product 
is  equal  to  0,  whatever  the  values  of  the  other  factors. 
Such  a  factor  is  called  a  zero  factor. 

Thus,  abed  =  0,  if  a,  6,  c,  or  d  =  0. 

22.  Coefficients.  Any  factor  of  a  product  may  be  con- 
sidered as  the  coefficient  of  the  remaining  factors ;  that  is, 
the  co-factor  of  the  remaining  factors.  Coefficients  expressed 
by  letters  are  called  literal  coefficients ;  expressed  by  Arabic 
numerals,  numerical  coefficients. 

Thus,  in  7  x,  7  is  the  numerical  coefficient  of  x ;  in  ax,  a  is  the 
literal  coefficient  of  x. 

If  no  numerical  coefficient  is  written,  1  is  understood. 


6  DEFINITIONS  AND  NOTATION. 

23.  Powers  and  Roots.  When  a  number  is  taken  a  num- 
ber of  times  as  a  factor,  the  result  is  called  a  power  of  the 
factor.  When  a  number  is  the  product  of  equal  factors, 
one  of  the  equal  factors  is  called  a  root  of  the  number. 

24.  Indices  or  Exponents  of  Powers.  An  index  or  exponent 
of  a  power  is  a  number-symbol  written  at  the  right  of  and 
a  little  above  the  number. 

If  the  exponent  is  a  whole  number,  it  shows  the  number 
of  times  the  given  number  is  taken  as  a  factor. 

Thus,  a1,  or  simply  a,  denotes  that  a  is  taken  once  as  a  factor ;  a2 
denotes  that  a  is  taken  twice  as  a  factor ;  a8  denotes  that  a  is  taken 
three  times  as  a  factor ;  and  so  on.  These  are  read  :  the  first  power 
of  a ;  the  second  power  of  a ;  the  third  power  of  a ;  and  so  on.  We 
write  a3  for  aaa,  a4  for  aaaa,  an  for  aaaa to  n  factors. 

Note.  The  second  power  of  a  number  is  often  called  the  square  of 
that  number ;  thus,  a2  is  called  the  square  of  a,  because  if  a  denotes 
the  number  of  units  of  length  in  the  side  of  a  square,  a2  denotes  the 
number  of  units  of  surface  in  the  square.  The  third  power  of  a 
number  is  often  called  the  cube  of  that  number ;  thus,  a3  is  called  the 
cube  of  a,  because  if  a  denotes  the  number  of  units  of  length  in  the 
edge  of  a  cube,  a8  denotes  the  number  of  units  of  volume  in  the  cube. 

25.  The  meaning  of  coefficient  and  exponent  must  be 
carefully  distinguished.     Thus, 

4a  =  a  +  a  +  a  +  a; 
a*  =  aXaX.aXa. 
Ifa  =  3,  4a  =  3  +  3  +  3  +  3  =  12. 

a*  =  3  X  3  X  3  x  3  =  81. 

26.  Indices  of  Roots.  An  index  of  a  root  is  a  number- 
symbol  written  above  the  radical  sign  to  indicate  the  re- 
quired root. 

Thus,  Va,  or  simply  Va,  means  one  of  the  two  equal  factors  of  a, 

that  is,  the  square  root  of  a ;  Va  means  one  of  the  three  equal  factors 
of  a,  that  is,  the  cube  root  of  a ;  and  so  on. 


DEFINITIONS  AND  NOTATION.  7 

Algebraic  Expressions. 

27.  An  Algebraic  Expression.  An  algebraic  expression  is 
a  number  written  with  algebraic  symbols.  An  algebraic 
expression  may  consist  of  one  symbol,  or  of  several  symbols 
connected  by  signs. 

Thus,  a,  3  a&c,  5  a  +  2  &  —  3  c,  are  algebraic  expressions. 

28.  Terms.  A  term  is  an  algebraic  expression  of  one  sym- 
bol, or  of  several  symbols  not  separated  by  the  sign  +  or  — . 

Thus,  a,  5  xy,  2  ab  X  4  cd,  j—z  are  algebraic  expressions  of  one  term 
each.     A  term  may  be  separated  into  parts  by  the  sign  Xor-r. 

29.  Similar  Terms.  If  terms  have  the  same  letters,  and 
each  letter  has  the  same  exponent  in  all  the  terms,  they  are 
called  similar  terms  or  like  terms. 

Thus,  3x2y3,  5x2y3,  and  7  x2y*  are  similar  terms. 

30.  Simple  Expressions.  An  algebraic  expression  of  one 
term  is  called  a  simple  expression  or  a  monomial. 

Thus,  5  xy,  7  a  X  2  6,  7  a  -r  2  6,  are  simple  expressions. 

31.  Compound  Expressions.  An  algebraic  expression  of 
two  or  more  terms  is  called  a  compound  expression  or  a  poly- 
nomial. 

Thus,  5  xy  +  7  a,  2  x  —  y  —  3  z,  are  compound  expressions. 

32.  A  polynomial  of  two  terms  is  called  a  binomial ;  of 
three  terms,  a  trinomial.  % 

Thus,  3 a  —  b  is  a  binomial ;  and  3a  —  6  +  cisa  trinomial. 

33.  Plus  and  Minus  Terms.  A  term  preceded  by  the 
sign  -f  is  called  a  plus  term ;  and  a  term  preceded  by  the 
sign  —  is  called  a  minus  term.  The  sign  +  before  a  single 
term,  and  before  the  first  of  a  series  of  terms  is  omitted. 


8  DEFINITIONS  AND  NOTATION. 

34.  A  plus  term  and  a  minus  term  cancel  each  other  when 
combined,  if  both  terms  stand  for  the  same  number. 

35.  The  Numerical  Value  of  an  Expression.  The  result 
obtained  by  putting  particular  values  for  the  letters  of  an 
expression  and  performing  the  indicated  operations  is  called 
the  numerical  value  of  the  expression. 

Numerical  Values  of  Simple  Expressions. 

1.  If  a  =  3,  find  the  numerical  values  of  4  a  and  a*. 

4a  =  4Xa  =  4x3  =  12; 
and  a4  =  aXaXaXa  =  3X3x3x3  =  81. 

2.  If  a  =  5,  b  =  6,  c  =  7,  find  the  numerical  value  of  the 
expression  J^  abc. 

T9¥a6c  =  ^X5x6x7  =  135. 

3.  If  x  =  2,  y  =  3,  find  the  numerical  value  of  5  x V. 

5  xV  =  5X22X38  =  5X4X27  =  540. 

4.  If  x  =  4,  y  =  6,  find  the  numerical  value  of  f  x*y. 

\&y  =  *  X  42  X  6  =  f  X  16  X  6  =  64. 

5.  If  x  =  2,  y  =  3,  z  =  4,  find  the  numerical  value  of 
$x?y  +  3z*. 

8z*y  _  8  X2X2X3_1 
3  z»       3X4X4X4~2* 

6.  If  x  =  3,  find  the  numerical  values  of  ^/Ax2;  Via?; 
and  2  V(  9  a;2)- 

V4x2  =  V4  x  32  =  2  x  9  =  18. 

Viz2  =  V4  X  32  =  V§6    =6. 

2  V(9x2)  =  2  V(9  X  32)  =  2  X  9  =  18. 

Note.  When  no  vinculum  or  parenthesis  is  used,  a  radical  sign 
affects  only  the  symbol  immediately  following  it. 


1. 

15*. 

11. 

f&V. 

2. 

3ab. 

12. 

*«y. 

3. 

7  by. 

13. 

w. 

4. 

5bd. 

14. 

§V- 

5. 

9y2. 

15. 

f  x  V. 

6. 

3b2c. 

16. 

tv<w. 

7. 

±c2x2. 

17. 

A«*V. 

8. 

2b6x. 

18. 

f  a4z2. 

9. 

b2cy2. 

19. 

*&y. 

10. 

a4b3c2. 

20. 

i«w 

3ft¥ 

26. 

V*2/. 

8  6c 

27. 

■y/dx. 

U2y 

28. 

!/Se. 

2dc 

29. 

^/bdx8. 

x2y2 

30. 

V(%). 

5b2c2 

31. 

^Jab2x2. 

x2y2z2 

32. 

■^Jxhfz3. 

3c3d? 

33. 

Zj3be2d. 

10  a5b5 

34. 

2V^W. 

oV>ba 

35. 

c^/dx2. 

DEFINITIONS  AND  NOTATION. 


Exercise  1. 

If  a  =  1,  b  =  2,  c  =  3,  d  =  4:,  x  =  5,  y  =  6,  z  =  0,  find 
the  numerical  value  of : 


21 


22. 


23. 


24. 


25. 


Numerical  Values  of  Compound  Expressions. 

36.  Each  term  should  be  written  in  the  algebraic  form 
by  omitting  the  sign  X  between  a  numerical  factor  and  a 
literal  factor  or  between  two  literal  factors.  The  opera- 
tions indicated  in  a  term  must  be  performed  before  the 
operation  indicated  by  the  sign  prefixed  to  the  term. 

37.  The  parts  of  a  term  are  combined  in  the  order  of  the 
signs  X  and  ■**  from  left  to  right. 

The  terms  of  an  expression  are  combined  in  the  order  of 
the  signs  +  and  —  from  left  to  right. 

Thus,  60  -  40  -i-  5  X  3  -  20  =  60  -  ^  X  3  -  20  =  16. 

38.  The  terms  may  be  arranged  in  any  order  before 
combining  them.  This  is  called  the  commutative  law  for 
addition  and  subtraction. 


10  DEFINITIONS  AND  NOTATION. 


Numerical  Values  of  Compound  Expressions. 

1.  If  b  =  10,  c  =  2,  y  =  5,  find  the  numerical  value  of 
66-  (8y  +  2c)c-2cy. 

66  —  (8y  -r  2 c)  c  —  2 cy  =  6  X  10  —  -4-0  X  2  —  2  X  2  X  5 
=  60  —  20  —  20  =  20. 

2.  If  x  =  7,  y  =  5,  find  the  numerical  value  of  • 

x  +  y 


(x  +  y)(x  —  y)  + 


«  — y 


C^y)(*-y)  +  ~|  =  (7  +  6)(7--6)  +  |i| 

=  12  X  2  +  V-  =  30. 


Exercise  2. 
If  a  =  1,  6  =  2,  c  =  3,  find  the  value  of : 

1.  7  a  —  be.  b.2a  —  b  +  c.         9.    V4  a^  +  2  c. 

2.  ac  +  6.  6.    a6  +  6c  —  ac.     10.    V6abc  —  2b. 

3.  4a&-c.  7.    62  +  a2  +  c2.       11.    c3  -  63. 

4.  6a6-6-c.        8.    5  6c2-2a6.        12.    -^c2  -  a\ 

13.    26  + 3c -a.  16.    6b  -  lObc  -*-  12a  4-  2c. 

14    <>  +  6)2-h2(c-a)3.       17.    5  c  h-  (6-  a)  -6  -a. 
15.    V66c"  -  6  -  c.  18.    -vW  -  VbW. 

If  a  =  1,  b  =  2,  c  =  3,  d  =  0,  find  the  value  of : 

19.  la  -be  +  6d.  25.  ^4abcd  +  2b*. 

20.  ac  +  6  —  d.  26.  -\/Tabcd  +  6C. 

21.  ±ab-cd-d.  27.  63  -  c  +  dc. 

22.  2a-H^.  28.  3c*-262  +  a. 

23.  ab  +  bc-ad.  29.  26  +(5c  -  3)-h(2c  -  a). 

24.  2a6-56c£  30.  3c2  -  2a9-  -  26b. 


DEFINITIONS  AND  NOTATION.  H 

Parentheses. 

39.  A  parenthesis  preceded  by  the  sign  -[-.  If  a  man  has 
10  dollars  and  afterwards  collects  3  dollars  and  then  2 
dollars,  it  makes  no  difference  whether  he  puts  the  3 
dollars  and  the  2  dollars  together  and  adds  their  sum  to 
his  10  dollars,  or  adds  the  3  dollars  to  his  10  dollars,  and 
then  the  2  dollars. 

The  first  process  is  represented  byl0-f(34-2). 
The  second  process  is  represented  by  10  +  3  4-  2. 

Hence,  10  +  (3  +  2)  =  10  +  3  +  2.  (1) 

If  a  man  has  10  dollars  and  afterwards  collects  3  dollars 
and  then  pays  a  bill  of  2  dollars,  it  makes  no  difference 
whether  he  pays  the  2  dollars  from  the  3  dollars  collected 
and  adds  the  remainder  to  his  10  dollars,  or  adds  the  3 
dollars  collected  to  his  10  dollars  and  pays  from  this  sum 
his  bill  of  2  dollars. 

The  first  process  is  represented  by  10  +  (3  —  2). 
The  second  process  is  represented  by  10  4-  3  —  2. 

Hence,  10  +  (3  -  2)  =  10  4- 3  -  2.  (2) 

If  we  use  general  symbols  in  (1)  and  (2),  we  have, 

«  +  (J  +  c)  =  a  +  J  +  c, 
and  a  4-  (b  —  c)  —  a  +  b  —  c.  Hence, 

We  have  the  general  rule  for  a  parenthesis  preceded  by  4~  * 

If  an  expression  within  a  parenthesis  is  preceded  by  the 
sign  +,  the  parenthesis  may  be  removed  without  making  any 
change  in  the  signs  of  the  terms  of  the  expression. 

Instead  of  a  parenthesis,  any  other  sign  of  aggregation 
may  be  used  and  the  same  rule  will  apply. 


12  DEFINITIONS  AND  NOTATION 

40.  A  parenthesis  preceded  by  the  sign  — .  If  a  man  with 
10  dollars  has  to  pay  two  bills,  one  of  3  dollars  and  one  of 
2  dollars,  it  makes  no  difference  whether  he  takes  3  dollars 
and  2  dollars  at  one  time,  or  takes  3  dollars  and  2  dollars 
in  succession,  from  his  10  dollars. 

The  first  process  is  represented  by  10  —  (3  +  2). 
The  second  process  is  represented  by  10  —  3  —  2. 

Hence,  10  -  (3  +  2)  =  10  -  3  -  2.  (1) 

If  a  man  has  10  dollars  consisting  of  two  5-dollar  bills, 
and  has  a  debt  of  3  dollars  to  pay,  he  can  pay  his  debt  by 
giving  a  5-dollar  bill  and  receiving  2  dollars. 

This  process  is  represented  by  10  —  5  +  2. 

Since  the  debt  paid  is  3  dollars,  that  is,  (5  —  2)  dollars, 
the  number  of  dollars  he  has  left  can  be  expressed  by 

10 -(5 -2). 
Hence,  10  -  (5  -  2)  =  10  -  5  +  2.  (2) 

If  we  use  general  symbols  in  (1)  and  (2),  we  have, 
a  —  (b  -f-  c)  =  a  —  b  —  c, 
and  a  —  (b  —  c)  —  a  —  b  +  c.  Hence, 

We  have  the  general  rule  for  a  parenthesis  preceded  by  — : 

If  an  expression  within  a  parenthesis  is  preceded  by  the 
sign  — ,  the  parenthesis  may  be  removed,  provided  the  sign 
before  each  term  within  the  parenthesis  is  changed,  the 
sign  -\-to  —  and  the  sign  —  to  +. 

Note.  If  the  vinculum  is  used,  the  sign  prefixed  to  the  first  term 
under  the  vinculum  must  he  understood  as  the  sign  before  the 
vinculum. 


Thus,     a  +  b  —  c  has  the  same  meaning  as  a  +  (b  —  c), 
and  a  —  b  —  c  has  the  same  meaning  as  a  —  (b  —  c). 


DEFINITIONS  AND  NOTATION.  13 

♦ 
Exercise  3. 

Eemove  the  parentheses  and  combine : 

1.  9  +  (3  +  2).     5.  9  -  (8  -  6).       9.  (3  -  2)  -  (2  -  1). 

2.  9  +  (3  -  2).     6.  8  -  (7  -  5).     10.  (7  -  3)  -  (3  -  2). 

3.  7  +  (5  +  1).     7.  9  -  (6  +  1).     ll.  (8  -  2)  -  (5  -  3). 

4.  7  +  (5  -  1).     8.  8  -  (3  +  2).     12.  15  -  (10  -  3  -  2). 

If  a  =  10,  b  =  5,  c  =  4,  d  =  2,  find  the  value  of: 

13.  (a  +  6)  +  (c  +  d).  15.    (a-b)-(c-d). 

14.  (a  +  &)-(<?-<*).  16.    (a -$)  +  («  —  d). 

Product  of  a  Compound  by  a  Simple  Factor. 

41.  In  finding  the  product  of  4  (5  +  3),  it  makes  no  dif- 
ference in  the  result  whether  we  multiply  the  sum  of  5  and 
3  by  4,  or  multiply  5  by  4  and  3  by  4  and  add  the  products. 

By  the  first  process,  4(5  +  3)  =  4x8  =  32. 

By  the  second  process,  4  (5  +  3)  =  (4  X  5  +  4  X  3)  =  32. 

In  like  manner,  4  (5  —  3)  =  4  X  2  =    8, 

and  4(5-3)  =  (4x  5-4x3)=    8, 

In  general  symbols,  a  (b  +  c)  =  ab  -f  ac, 
and  a(b  —  c)  =  ab  —  ac. 

This  is  called  the  distributive  law  for  multiplication. 

42.  The  order  of  the  factors  is  immaterial. 
Thus,  4(5  +  3)^4x5  +  4x3  =  32,       ' 

and  (5  +  3)  4  =  5  X  4  +  3  X  4  =  32. 

In  general  symbols,         ab  =  ba. 
This  is  called  the  commutative  law  for  multiplication. 
Perform  the  indicated  operations : 

J.    x  +  3(a-b).  2.    x-S(a-b). 

1.  x  +  3  (a  —  b)  =  x  +  (3a  -  3  b)  =  x  +  3  a  -  36. 

2.  x  —  3(a-&)  =  z-(3a-36)  =  x  —  3a +  36. 


14  DEFINITIONS  AND  NOTATION. 

Exercise  4. 

Perform  the  indicated  operations,  and  find  the  numerical 
value  of  each  expression,  if  a  =  5,  b  =  4,  c  =  3 : 

1.  3(ab  +  c).       7.  bc  +  a(b  —  c).       13.   (a-c)b-a. 

2.  4:(ac  +  b).       8.  b  +  2(a-c).        14.  (a -&)<*  + 2c. 

3.  2(aft  — c).       9.  ac-2(6  +  c).       15.  (6  +  c)c-2&. 

4.  5(ac-&).      10.  5ac-2(b2  +  b).     16.  (a-c)7-3e. 

5.  7(Jc-a).      11.  2b±3(a-c).     17.   <>-c)c2-2c2. 

6.  ac(b-c).      12.  a6  +  &(£-c).       18.   (a2  — &2)c-a2. 

Quotient  of  a  Compound  by  a  Simple  Expression. 

43.  In  finding  the  quotient  of  (8  +  4)^-2  it  makes  no 
difference  in  the  result  whether  we  divide  the  sum  of  8  and 
4  by  2,  or  divide  8  by  2  and  4  by  2,  and  add  the  quotients. 

By  the  first  process,  (8  +  4)  -*-  2  =  12  -h  2  =  6. 

By  the  second  process,  (8  +  4)  -s-  2  =  (8-^2  +  4-^  2)  =  6. 

In  general  symbols, =  --}--, 

,  a  —  b      a      b 

and  = 

c  c      c 

This  is  called  the  distributive  law  for  division. 

Perform  the  indicated  operations : 

1.  x  +  (3a  +  3 b)  -J- 3.  2.   x-(3a  +  3b)+3. 

1.  x  +  (3a  +  36)  -f  3  =  »  +  (a  +  6)  =  x  +  a  +  6. 

2.  *  —  (3  a  +  3  fe\  -fr  3  =  x  —  (a  +  b)  =  x  —  a  —  b. 

Exercise  5. 

Perform  the  indicated  operations,  and  find  the  numerical 
value  of  each  expression,  if  a  =  8,  b  =  4,  c  =  2 : 


1.    (5  +  c)-s-c.       5.    O^  +  c) 


(a  +  c)  +  b.       6.    (ac +  &)  +  £.       10.    (62  -  c2)  -5-  J. 


3.    (a-b)  +  b.      7.    (ac  -  &) 


c.  9.    (£2  +  c2) 


&.       11.    (a2-c2)H-c2. 


4.    (6-c)-5-c.       8.    (oi-o)-HO.       12.    (a1- &■)■+■  P. 


CHAPTER  IL 
SIMPLE  EQUATIONS. 

44.  Equations.  An  equation  is  a  statement  in  symbols 
that  two  expressions  stand  for  the  same  number. 

Thus,  the  equation  Sx  +  2  =  8  states  that  Sx  +  2  and  8  stand  for 
the  same  number. 

45.  That  part  of  the  equation  which  precedes  the  sign 
of  equality  is  called  the  first  member,  or  left  side,  and  that 
part  of  the  equation  which  follows  the  sign  of  equality  is 
called  the  second  member,  or  right  side. 

46.  An  equation  containing  letters,  if  true  for  all  values 
of  the  letters  involved,  is  called  an  identical  equation ;  but 
if  it  is  true  only  for  certain  particular  values  of  the  letters 
involved,  it  is  called  an  equation  of  condition. 

Thus,  a  +  b  =  b  +  a,  which  is  true  for  all  values  of  a  and  b,  is  an 
identical  equation ;  and  3  x  +  2  =  8,  which  is  true  only  when  x  stands 
for  2,  is  an  equation  of  condition. 

For  brevity,  an  identical  equation  is  called  an  identity, 
and  an  equation  of  condition  is  called  simply  an  equation. 

47.  We  often  employ  an  equation  to  discover  an  unknown 
number  from  its  relation  to  known  numbers.  We  usually 
represent  the  unknown  number  by  one  of  the  last  letters  of 
the  alphabet,  as  x,  y,  z ;  and  the  known  numbers  by  the 
first  letters,  a,  b,  c,  and  by  the  Arabic  numerals. 

48.  Simple  Equations.  Equations  which,  when  reduced 
to  their    simplest   form,    contain    only    the   first   power 


16  SIMPLE  EQUATIONS. 

of  the  unknown  numbers  are  called  simple  equations,  or 
equations  of  the  first  degree. 

Thus,  7x  +  5  =  4x  +  14,  and  ax  +  b  =  c  are  simple  equations  in  x. 

49.  Two  or  more  like  terms  may  be  combined  to  form  a 
single  like  term  by  uniting  their  numerical  coefficients. 

Thus,  3 ax  +  ax  =  4 ax ;  and  box  —  Sax  =  2 ax. 

50.  To  Solve  an  Equation  with  One  Unknown  Number  is  to 
find  the  unknown  number;  that  is,  to  find  the  number 
which,  when  substituted  for  its  symbol  in  the  given  equa- 
tion, renders  the  equation  an  identity. 

This  number  is  said  to  satisfy  the  equation,  and  is  called 
the  root  of  the  equation. 

51.  Axioms.  In  solving  an  equation,  we  make  use  of 
the  following  self-evident  truths,  called  axioms : 

Ax.  1.  If  equal  numbers  are  added  to  equal  numbers,  the 
sums  are  equal. 

Ax.  2.  If  equal  numbers  are  subtracted  from  equal  num- 
bers, the  remainders  are  equal. 

Ax.  3.  If  equal  numbers  are  multiplied  by  equal  num- 
bers, the  products  are  equal. 

Ax.  4.  If  equal  numbers  are  divided  by  equal  numbers, 
the  quotients  are  equal. 

Ax.  5.  If  two  numbers  are  equal  to  the  same  number, 
they  are  equal  to  each  other. 

52.  Transposition  of  Terms.  It  becomes  necessary  in 
solving  simple  equations  to  bring  all  the  terms  that  contain 
the  symbols  for  the  unknown  numbers  to  one  side  of  the 
equation,  and  all  the  other  terms  to  the  other  side.  This 
process  is  called  transposing  the  terms. 


SIMPLE  EQUATIONS.  17 

1.  Find  the  number  for  which  x  stands  when 

x  —  b  =  a. 
Add  b  to  each  side,  x  —  b  +  b  =  a  +  b.  (Ax.  1) 

Cancel  -b  +  5,  x  =  a  +  b.  (§34) 

The  result  is  the  same  as  if  we  had  transposed  —  b  from 
the  left  side  to  the  right  side  and  changed  its  sign. 

2.  Find  the  number  for  which  x  stands  when 

x  +  b  =  a. 
Subtract  b  from  each  side,  x  +  b  —  b  =  a  —  b.       (Ax.  2) 
Cancel  +  b~b,  x  =  a  -  b.        (§34) 

In  this  case,  we  have  transposed  b  from  the  left  side  to 
the  right  side  and  changed  its  sign. 

We  can  proceed  in  like  manner  in  any  other  case. 

Hence,  the  general  rule : 

53.  Any  term  may  be  transposed  from  one  side  of  an 
equation  to  the  other,  provided  its  sign  is  changed. 

It  follows  from  axioms  1  and  2  that : 

54.  Any  term  that  occurs  with  the  same  sign  on  both  sides 
of  an  equation  may  be  cancelled. 

If  we  transpose  each  term  of  the  equation, 

c  —  x  =  a  —  b,  (1) 

we  have  b  —  a  =  x  —  c. 

That  is,  x  —  c  =  b  —  a.  (2) 

Equation  (2)  is  the  same  as  (1)  with  the  sign  before 
each  term  changed.     Hence : 

55.  The  sign  of  every  term  of  an  equation  may  be  changed 
without  destroying  the  equality. 


18  SIMPLE  EQUATIONS. 

56.  Numerical  Equations.  An  equation  in  which  all  the 
known  numbers  are  expressed  by  Arabic  numerals  is  called 
a  numerical  equation. 

Solution  of  Simple  Numerical  Equations  in  X 

1.    Solve  Sx-7  =  14:-Ax. 

Transpose  —  4  x  to  the  left  side  and  —  7  to  the  right  side.. 


3z  +  4x  =  14  +  7. 

(§58) 

Combine, 

7x  =  21. 

(§  49) 

Divide  by  7, 

x  =  S. 

(Ax.  4) 

2.    Solve  the  equation 

1  -  4  (x  -  2)  =  Ix  -  3  (3  x  -  1). 
Multiply  the  compound  factor  by  the  simple  factor  in  each  side, 

1  —  (4  x  —  8)  =  7  x  —  (9  x  —  3). 
Remove  the  parenthesis  in  each  side, 

l-4z  +  8  =  7a;-9x  +  3.  (§40) 

Transpose,        9x  —  4ce  —  7jc  =  3  —  1  —  8. 
Change  the  signs  of  all  the  terms, 

4z  +  7x  —  9z  =  l  +  8-3.  ^§55) 

Combine,  2  x  =  6.  (§  49) 

Divide  by  2,  x  =  3.  (Ax.  4) 

57.  To  Solve  a  Simple  Numerical  Equation  in  ^r,  therefore : 
Transpose  all  the  terms  that  contain  x  to  the  left  side,  and 

all  the  other  terms  to  the  right  side.     Combine  similar  terms, 
and  divide  both  sides  by  the  coefficient  of  x. 

58.  Verification.  If  the  value  found  for  x  is  substituted 
for  x  in  the  original  equation,  and  the  equation  reduces  to 
an  identity,  the  value  of  x,  that  is,  the  root  of  the  equation, 
is  said  to  be  verified. 

Note.  In  verifying  a  solution,  as  in  solving  an  equation,  it  is  im- 
portant to  notice  that  the  signs  of  all  the  terms  may  be  changed. 


SIMPLE  EQUATIONS. 

Show  that  x  stands  for  3  in  the  equation 

3a;-7  =  14-4a;. 
Put  3  for  x  in  this  equation,  and  we  have 

3x3-7  =  14 -4X3, 
or,  9  -  7  =  14  -  12, 

that  is,  2  =  2. 

Exercise  6. 
Find  the  value  of  x,  and  verify  the  answer : 

1.  5a; -4  =  16.  10.    14a;  -79  =  8a: -25. 

2.  3a; +  4  =  25.  11.    5a; -4  =  12- 3a\ 

3.  24a; -7a;  =  34.  12.    Ix  +  4  =  3x  +  24. 

4.  16a;  =  7a; +  81.  13.    12a  -  16  =  8  +  6x. 

5.  3a;  =  55 -2a;.  14.    4a; -10  =  14  + 2a;. 

6.  5  a;  =  3  a; +  6.  15.    2 x  —  5  =  7  —  x. 

7.  7  a;  =  6a; +  4.  16.    4a; -14  =  a; -2. 

8.  5x  =  28-2x.  17.    4a;  -  11  =  2a;  -  5. 

9.  2a;  =  11+ a-.  18.    4a;  -  10  =  3a;  -  5. 

19.  5  (x  +  1)  +  6  (x  +  2)  =  7  (a  4-  3). 

20.  4  (a;  +  7)  -  36  =  13  (x  -  2). 

21.  6  (3 aj  -  1)  -8a;  =  140  +  2  (x  -  1). 

22.  3  (3x  -  2)  -  6  (4  -  x)  =  24a;  -4  (7a;  -2). 

23.  3  (aj  +  13)  -  15  =  4  (x  -  2)  -  9. 

24.  10a;  -  (x  -10)  =  3a;  +  52. 

25.  3a;  -  (x  +  5)  -  (x  -  3)  =  10  -  a;. 

26.  x2  +  8a;  -  (a;2  -  aj)  =  5  (x  +  3)  +  5. 

27.  5  -  x  +  4  (a;  -  1)  -  (a;  -  2)  =  15. 

28.  3  (a?  +  10)  +  4  (x  +  20)  +  5x  =  185  -  3a;. 

29.  2  (a?  -  2)  +  3  (a?  -  3)  +  4  (a?  -  4)  =  3a;  +  7. 

30.  (5a;  +  3)  -  2  (x  -  1)  +  (1  -  x)  =  4  (9  -  a;). 

31.  7  -  21  (x  +  3)  =  13  -  15  (2x  -  5). 

32.  5  (a?  -  3)  -  7  (6  -  x)  +  29  =  50  -  3  (8  -  x). 


& 


20  SIMPLE  EQUATIONS. 


Statement  and  Solution  of  Problems. 

59.  To  express  in  algebraic  language  the  conditions  of  a 
problem  that  are  stated  in  common  language  is  generally 
very  difficult  for  the  beginner.  We  will  therefore  give  an 
exercise  on  translating  common  language  into  algebraic 
language  before  proceeding  to  the  solutions  of  problems. 


Exercise  7. 

1.  Write  in  symbols  :  a  diminished  by  b ;  a  increased 
by  b  \  a  multiplied  by  b  ;  a  divided  by  b  ;  the  square  of  a ; 
the  square  root  of  a ;  the  cube  root  of  a ;  the  square  of  a 
multiplied  by  the  fourth  power  of  b. 

2.  If  a  man  walks  x  miles  an  hour,  how  many  miles 
will  he  walk  in  4  hours  ?    in  a  hours  ?     \  A 

3.  If  a  man  walks  3  miles  an  hour,  how  many  hours 
will  it  take  him  to  walk  12  miles  ?   x  miles  ?  f 

4.  If  a  man  walks  x  miles  an  hour,  how  many  hours 

will  it  take  him  to  walk  20  miles  ?   y  miles  ?    i>    Jk 

n        \ 

5.  What  is  the  divisor,  if  the  dividend  is  20  and  the 

quotient  5  ?    if  the  dividend  is  a  and  the  quotient  b  ?    *\  -*< 

6.  What  is  the  dividend,  if  the  divisor  is  4,  the  quotient 
3,  and  the  remainder  2  ?  if  the  divisor  is  d,  the  quotient 
q,  and  the  remainder  r  ?  \  ^    o*-fy  *t  ^ 

7.  What  is  the  quotient,  if  the  dividend  is  22,  the  divi- 
sor 4,  and  the  remainder  2  ?  if  the  dividend  is  p}  the  divi- 
sor d,  and  the  remainder  r?  $~      ^C^  ^ 

8.  What  is  the  divisor,  if  the  dividend  is  22,  the  quo- 
tient 4,  and  the  remainder  2  ?  if  the  dividend  is  p,  the 
quotient  q,  and  the  remainder  r  ?  5"    j_ 

9.  If  one  part  of  25  is  10,  what  is  the  other  part  ? 


SIMPLE  EQUATIONS.  21 

10.  If  one  part  of  30  is  x,  what  is  the  other  part  ? 

11.  If  one  part  of  x  is  c,  what  is  the  other  part  ?  '%^'(^0, 

12.  If  the  sum  of  two  numbers  is  40,  and  one  of  them  is 
25,  what  is  the  other  ?     '  T 

13.  If  the  sum  of  two  numbers  is  x,  and  one  of  them  is 
5,  what  is  the  other  ?     -fi  ^3"  -  >S 

14.  If  the  sum  of  two  numbers  is  s,  and  one  of  them  is 
a,  what  is  the  other  ?  f  ^  £. 

15.  If  the  difference  of  two  numbers  is  7,  and  the  smaller 
number  is  13,  what  is  the  greater  number  ?    *-0 

16.  If  the  difference  of  two  numbers  is  a,  and  the  smaller 
number  is  x,  what  is  the  greater  number  ?      °U  t-  pf 

17.  If  the  difference  of  two  numbers  is  c,  and  the  greater 
number  is  x,  what  is  the  smaller  number  ?     ^  -  O 

18.  Henry  is  a  years  old  to-day.  How  old  was  he  4 
years  ago  ?     How  old  will  he  be  in  4  years  ?  a  ~  ^ 

19.  John  is  x  years  old  to-day.  How  old  was  he  b  years 
ago  ?     How  old  will  he  be  a  years  hence  ?     "%*~  at       n  ^ ^ 

20.  By  how  much  does  5  x  exceed  3  x  ?      "*-  ~fi 

21.  By  how  much  does  x  exceed  a  ?     Vs  ^ 

22.  How  much  does  a  lack  of  being  x  ?    V  ~ 

23.  Write  the  excess  of  2x  -f  3  over  a?  +  1. 

Note.  If  the  number  to  be  subtracted  is  a  compound  expression 
it  must  be  enclosed  by  a  parenthesis.  Thus,  the  excess  of  2  x  +  3 
over  x  +  1  is  2  x  +  3  —  (x  +  1). 

24.  Write  the  excess  of  3a;  over  4  (18  —  x). 

25.  Write  the  excess  of  x  —  50  over  80  —  x. 

26.  What  is  the  excess  of  2  x  —  24  over  80  —  5  a;  ? 

27.  What  is  the  excess  of  5  a;  -f-  24  over  60  — x? 

28.  Express  in  cents  a  half-dollars  and  b  quarters. 


22  SIMPLE  EQUATIONS. 

29.  Express  in  cents  a  dollars  b  dimes  and  c  cents. 

30.  A  man  has  a  dollars.  If  lie  spends  b  half-dollars 
and  c  dimes,  how  many  cents  has  he  left  ?  JS"T> '{>-+-  j&c)  *  "k 

31.  A  man  has  x  dollars  y  dimes  and  z  cents.  If  he 
spends  a  half-dollars  and  b  quarters,  how  many  cents  has 
he  left?   /^  +-  /O^Z  -  (i«^  f  ^U]   -^^ 

32.  A  man  makes  a  journey  of  236  miles.  He  travels  a 
miles  by  train,  c  miles  by  boat,  and  the  remainder  on  foot. 
How  far  does  he  go  on  foot  ?   %~%$  (f  —  ^  ~  v 

33.  A  train  is  running  at  the  rate  of  a  miles  an  hour. 
How  many  miles  will  it  travel  in  m  hours  ?   c^^^ 

34.  The  floor  of  a  square  room  measures  a  feet  each 
way.  How  many  square  yards  q|  oilcloth  will  b&  required 
to  cover  it?    b  oj-    ^        "  J>  i  %^"  •<"* 

35.  The  floor  of  a  rectangular  room  measures  a  feet  by 
b  feet.  How  many  square  yards  of.  oilcloth  will  be  re- 
quired to  cover  it  ?  S^l^      _2^-    ■  J^  '  -~^T 

36.  A  rectangular  floor  is  a  feet  long  and  b  feet  wide. 
In  the  middle  of  the  floor  there  is  a  square  carpet  c  feet  on 
a  side.     How  many  square  yards  of  the  floor  are  bare  ? 

60.  In  stating  problems,  x  must  not  be  put  for  money, 
length,  time,  weight,  etc.,  but  for  the  required  number  of 
specified  units  of  money,  length,  time,  weight,  etc. 

Each  statement  must  be  made  in  algebraic  symbols,  and 
the  meaning  of  each  algebraic  statement  should  be  written 
out  in  full,  in  common  language. 

After  the  algebraic  statements  are  written,  it  is  necessary 
and  sufficient,  in  problems  involving  only  one  unknown 
number,  to  select  two  expressions  that  stand  for  the  same 
number,  and  to  make  them  the  members  of  the  required 
equation.     (Ax.  5.) 


SIMPLE  EQUATIONS.  23 

Problems  Stated  and  Solved. 

1.  Three  times  a  certain  number  is  equal  to  the  number 
increased  by  20.     Find  the  number. 

Let  x  =  the  number. 

Then  3  x  =  3  times  the  number ; 

and  x  +  20  =  the  number  increased  by  20. 

But  the  last  two  expressions  are  equal. 
Therefore,  3  x  —  x  +  20. 

Transposing,  3  x  —  x  =  20. 

Combining,  .  2  x  =  20. 

Dividing  by  2,  x  —  10. 

2.  John  has  three  times  as  many  oranges  as  James,  and 
they  together  have  32.     How  many  has  each  ? 

Let  x  stand  for  the  number  of  oranges  James  has. 

Then      3  x  is  the  number  of  oranges  John  has ; 
and      x  +  3  x  is  the  number  of  oranges  they  together  have. 
But  32  is  the  number  of  oranges  they  together  have. 
Therefore,  <c  +  3x  =  32. 

Combining,  4ze  =  32. 

Dividing  by  4,  x  =  8. 

Multiplying  by  3,  3  x  =  24. 

Therefore,  James  has  8  oranges,  and  John  has  24  oranges. 

Note.     Beginners  in  stating  the  preceding  problem  generally  write: 
Let  x  =  what  James  had. 

Now,  we  know  what  James  had.     He  had  oranges,  and  we  are  to 
discover  simply  the  number  of  oranges  he  had. 

3.  James  and  John  together  have  $24,  and  James  has 
$8  more  than  John.     How  many  dollars  has  each  ? 

Let  x  stand  for  the  number  of  dollars  John  has. 

Then  x  +  8  is  the  number  of  dollars  James  has ; 

and  x  +  (x  +  8)  is  the  number  of  dollars  they  together  have. 

But  24  is  the  number  of  dollars  they  together  have. 


24  SIMPLE  EQUATIONS. 

Therefore,  x  +  (x  +  8)  =  24. 

Kemoving  the  parenthesis,  x  +  x  +  8  =  24. 
Combining,  2  x  =  16. 

Dividing  by  2,  x  —  8. 

Adding  8  to  each  side,  x  +  8  =  16. 

Therefore,  John  has  $8,  and  James  has  $16. 

Note.     The  beginner  must  avoid  the  mistake  of  writing 
Let  x  =  John's  money. 

We  are  required  to  find  the  number  of  dollars  John  has,  and  there- 
fore x  must  represent  this  required  number. 

4.  The  sum  of  two  numbers  is  18,  and  three  times  the 
greater  number  exceeds  four  times  the  smaller  by  5.  Find 
the  numbers. 

Let  x  =  the  greater  number. 

Then,  since  18  is  the  sum,  and  x  is  one  of  the  numbers,  the  other 
number  must  be  the  sum  minus  x.     Hence, 

18  —  x  =  the  smaller  number. 
Now,  three  times  the  greater  number  is  3  x,  and  four  times  the  less 
number  is  4  (18  —  x);  and  3x  —  4  (18  —  x)  is  equal  to  the  excess  of 
three  times  the  greater  number  over  four  times  the  smaller  number. 
But  5  =  this  excess. 

.-.  3x-4(18-x)  =  6. 
.-.  3x  — (72  — 4x)  =  5, 
or         .  3x  —  72  +  4x  =  5. 

.-.  7x  =  77, 
and  x  =  11. 

Therefore,  the  numbers  are  11  and  7. 

5.  A  had  three  times  as  much  money  as  B.  He  gave 
B  $10,  and  then  had  only  twice  as  much  as  B.  How  much 
had  each  at  first  ? 

Let        x  =  the  number  of  dollars  B  had  at  first 
Then  3  x  =  the  number  of  dollars  A  had  at  first. 
Now,  3  x  —  10  is  the  number  of  dollars  A  had  after  giving  $10  to 
B.  and  x  +  10  is  the  number  of  dollars  B  then  had 


SIMPLE  EQUATIONS.  25 


Since  A's  money  is  twice  B's, 

3x  —  10  =  2(x  +  10) 
3x  —  10  =  2x  +  20 
3x-2x  =  20  +  10 
x  =  30 
3x  =  90. 

Therefore,  B  had  $30  and  A  had  $90. 


6.  A  has  $7  in  half-dollars  and  quarters.     If  he  has  24 
coins  in  all,  how  many  are  halves  and  how  many  quarters  ? 

Let  x  =  the  number  of  halves. 

Then  24  —  x  =  the  number  of  quarters. 

50  x  =  the  value  in  cents  of  the  halves. 
25  (24  —  x)  =  the  value  in  cetits  of  the  quarters. 
Therefore,  50  x  +  25  (24  —  x)  =  the  value  of  his  money  in  cents. 
But  700  =  the  value  of  his  money  in  cents. 

Hence,      50  x  +  25  (24  —  x)  =  700 
50x  +  600-25x  =  700 
25x  =  100 
x  =  4 
and  24  —  x  =  20. 

Therefore,  he  has  4  half-dollars  and  20  quarters. 

7.  A  man  is  now  twice  as  old  as  his  son ;  15  years  ago 
he  was  three  times  as  old.     Find  the  age  of  each. 

Let  x  =  the  number  of  years  in  the  son's  age. 

Then      2  x  =  the  number  of  years  in  the  father's  age. 

x  —  15  =  the  number  of  years  in  the  son's  age  15  years  ago. 
2  x  —  15  =  the  number  of  years  in  the  father's  age  15  years  ago. 
But  15  years  ago  3  times  the  son's  age  was  equal  to  the  father's  age. 

Therefore,  3  (x  -  15)  =  2  x  -  16 

3x-45  =  2x-15 
x  =  30 
2x  =  60. 

Therefore,  the  son  is  30  years  old  and  the  father  60 
years  oldo 


26  SIMPLE  EQUATIONS. 

Exercise  8. 

1.  If  a  number  is  multiplied  by  9,  the  product  is  810. 
Find  the  number. 

2.  If  the  sum  of  the  ages  of  a  father  and  son  is  56  years, 
and  the  father  is  7  times  as  old  as  the  son,  what  is  the  age 
of  each  ? 

3.  The  sum  of  two  numbers  is  161,  and  the  greater  is  6 
times  the  less.     Find  the  numbers. 

4.  A  tree  100  feet  high  was  broken  so  that  the  part 
broken  off  was  9  times  the  length  of  the  part  left  standing. 
Find  the  length  of  each  part. 

5.  The  difference  of  two  numbers  is  7,  and  their  sum  is 
63.     Find  the  numbers. 

6.  The  difference  of  two  numbers  is  13,  and  their  sum  is 
59.     Find  the  numbers. 

7.  Divide  36  into  two  parts  so  that  one  part  shall  be 
greater  by  6  than  the  other  part. 

8.  Three  times  a  given  number  is  equal  to  the  number 
increased  by  36.     Find  the  number. 

9.  Three  times  a  given  number  diminished  by  20  is 
equal  to  the  given  number.     Find  the  number. 

10.  One  number  is  4  times  another,  and  their  difference 
is  24.     Find  the  numbers. 

11.  The  sum  of  two  numbers  is  48,  and  one  of  them 
exceeds  the  other  by  6.     Find  the  numbers. 

12.  The  sum  of  two  numbers  is  42,  and  5  times  the 
smaller  number  is  equal  to  the  larger  number.  Find  the 
numbers. 

13.  Find  three  consecutive  numbers,  x,  x  +  1,  and  x  +  2, 
whose  sum  is  108. 


SIMPLE  EQUATIONS.  27 

14.  Find  five  consecutive  numbers  whose  sum  is  70. 

15.  A  man  walks  4  miles  an  hour  for  x  hours,  and 
another  man  walks  3  miles  an  hour  for  a;  +  4  hours.  If 
they  each  walk  the  same  distance,  how  many  miles  does 
each  walk? 

16.  A  farmer  employed  two  men  to  build  112  rods  of 
wall.  One  of  them  built  on  the  average  4  rods  a  day,  and 
the  other  3  rods  a  day.     How  many  days  did  they  work  ? 

17.  Two  men  start  from  the  same  place  and  travel  in 
opposite  directions,  one  30  miles  a  day,  and  the  other  20 
miles  a  day.    In  how  many  days  will  they  be  350  miles  apart  ? 

18.  Two  men  start  from  the  same  place  and  travel  in  the 
same  direction,  one  30  miles  a  day,  and  the  other  20  miles 
a  day.     In  how  many  days  will  they  be  350  miles  apart  ? 

19.  A  man  bought  3  equal  lots  of  hay  for  $255.  For 
the  first  lot  he  gave  $17  a  ton,  for  the  second  $16,  for  the 
third  $18.     How  many  tons  of  each  kind  did  he  buy  ? 

20.  A  farmer  sold  a  quantity  of  wood  for  $84,  one  half 
of  it  at  $3  a  cord,  and  the  other  half  at  $4  a  cord.  How 
many  cords  of  each  kind  did  he  sell  ? . 

21.  If  2x  —  S  stands  for  37,  for  what  number  will 
7  +  x  stand  ? 

22.  At  an  election  two  opposing  candidates  received 
together  2000  votes,  and  one  received  100  more  votes  than 
the  other.     How  many  votes  did  each  candidate  receive  ? 

23.  If  a  number  is  multiplied  by  17,  the  product  is  136. 
Find  the  number. 

24.  The  sum  of  two  numbers  is  54,  and  the  greater  is 
seventeen  times  the  smaller  number.     Find  the  numbers. 

25.  A  tank  holding  1500  gallons  has  three  pipes.  The 
first  lets  in  8  gallons  a  minute,  the  second  10  gallons,  and 


28  SIMPLE  EQUATIONS. 

the  third  12  gallons  a  minute.     In  how  many  minutes  will 
the  tank  be  filled  ? 

26.  The  fore  and  hind  wheels  of  a  carriage  are  10  feet 
and  12  feet  respectively  in  circumference.  How  many 
feet  will  the  carriage  have  passed  over  when  the  fore  wheel 
has  made  100  revolutions  more  than  the  hind  wheel  ? 

27.  Divide  a  yard  of  tape  into  two  parts  so  that  one  part 
shall  be  6  inches  longer  than  the  other  part. 

28.  Divide  23  into  two  parts  such  that  the  sum  of  twice 
the  greater  and  three  times  the  smaller  is  57. 

29.  Four  times  the  smaller  of  two  numbers  is  three  times 
the  greater,  and  their  sum  is  63.     Find  the  numbers. 

30.  A  farmer  sold  a  sheep,  a  cow,  and  a  horse  for  $216. 
He  sold  the  cow  for  seven  times  as  much  as  the  sheep,  and 
the  horse  for  four  times  as  much  as  the  cow.  How  much 
did  he  get  for  each  ? 

31.  Distribute  $15  among  Thomas,  Richard,  and  Henry 
so  that  Thomas  and  Richard  shall  each  have  twice  as  much 
as  Henry. 

32.  Three  men,  A,  B,  and  C,  pay  $1000  taxes.  B  pays 
four  times  as  much  as  A,  and  C  pays  as  much  as  A  and  B 
together.     How  much  does  each  pay  ? 

33.  John's  age  is  three  times  the  age  of  James,  and  their 
ages  together  are  16  years.     What  is  the  age  of  each  ? 

34.  Twice  a  certain  number  increased  by  8  is  40.  Find 
the  number. 

35.  Three  times  a  certain  number  is  46  more  than  the 
number  itself.     Find  the  number. 

36.  One  number  is  four  times  as  large  as  another.  If  I 
take  the  smaller  from  12  and  the  greater  from  21  the 
remainders  are  equal.     What  are  the  numbers  ? 


SIMPLE  EQUATIONS.  29 

37.  Thirty  yards  of  cloth  and  20  yards  of  silk  together 
cost  $70 ;  and  the  silk  costs  twice  as  much  per  yard  as 
the  cloth.     How  much  does  each  cost  per  yard  ? 

38.  In  a  company  of  180  persons,  composed  of  men, 
women,  and  children,  there  are  twice  as  many  men  as 
women,  and  three  times  as  many  women  as  children. 
How  many  are  there  of  each  ? 

39.  Two  trains  traveling,  one  at  25  and  the  other  at  30 
miles  an  hour,  start  at  the  same  time  from  two  places  330 
miles  apart,  and  move  toward  each  other.  In  how  many 
hours  will  the  trains  meet  ? 

40.  Twelve  persons  subscribed  'for  a  new  boat,  but  two 
being  unable  to  pay,  each  of  the  others  had  to  pay  $4  more 
than  his  share.     Find  the  cost  of  the  boat. 

41.  A  tree  84  feet  high  was  broken  so  that  the  part 
broken  off  was  five  times  the  length  of  the  part  left  stand- 
ing.    Required  the  length  of  each  part. 

42.  At  an  election  there  were  two  candidates,  and  2800 
votes  were  cast.  The  successful  candidate  had  a  majority 
of  160.     How  many  votes  were  cast  for  each  ? 

43.  Divide  20  into  two  parts  such  that  four  times  the 
greater  exceeds  three  times  the  smaller  by  17. 

44.  The  sum  of  two  numbers  is  50,  and  seven  times  the 
smaller  number  exceeds  three  times  the  greater  number  by 
10.     Find  the  numbers. 

45.  Divide  19  into  two  such  parts  that  twice  the  smaller 
part  exceeds  the  greater  by  two. 

46.  Three  times  the  excess  of  a  certain  number  over  6  is 
equal  to  the  number  plus  18.     Find  the  number. 

47.  Thirty-one  times  a  number  exceeds  80  by  as  much  as 
nine  times  the  number  is  less  than  80.     Find  the  number. 


30  SIMPLE  EQUATIONS. 

48.  Find  the  number  whose  double  diminished  by  3  ex- 
ceeds 80  by  as  much  as  the  number  itself  is  less  than  100. 

49.  Divide  19  into  two  parts  such  that  the  greater  part 
exceeds  twice  the  smaller  part  by  1  less  than  twice  the 
smaller  part.  , 

50.  A  man  is  now  twice  as  old  as  his  son ;  20  years  ago 
he  was  four  times  as  old  as  his  son.     Find  the  age  of  each. 

51.  A  man  is  four  times  as  old  as  his  son ;  in  20  years 
he  will  be  only  twice  as  old.     Find  the  age  of  each. 

52.  A  man  was  four  times  as  old  as  his  son  7  years  ago, 
and  will  be  only  twice  .as  old  as  his  son  7  years  hence. 
Find  the  age  of  each. 

63.  A  man  has  8  hours  for  an  excursion.  How  far  can 
he  ride  into  the  country  in  a  carriage  that  goes  at  the  rate 
of  9  miles  an  hour  so  as  to  return  in  time,  walking  back  at 
the  rate  of  3  miles  an  hour  ? 

54.  A  man  was  hired  for  26  days.  Every  day  he  worked 
he  was  to  receive  $3,  and  every  day  he  was  idle  he  was  to 
pay  $1  for  his  board.  At  the  end  of  the  time  he  received 
$62.     How  many  days  did  he  work  ? 

55.  A,  walking  4  miles  an  hour,  starts  two  hours  after  B, 
who  walks  3  miles  an  hour.  How  many  miles  must  A  walk 
to  overtake  B  ? 

56.  A  river  runs  1  mile  an  hour.  A  man  swims  a  certain 
distance  up  the  river  in  3  hours,  and  the  same  distance  down 
in  1  hour.     Find  his  rate  of  swimming  in  still  water. 

57.  A  man  bought  12  yards  of  velvet.  If  he  had  bought 
1  yard  iess  for  the  same  money,  each  yard  would  have  cost 
$1  more.     What  did  the  velvet  cost  a  yard  ? 

58.  A  and  B  have  together  $8;  A  and  C,  $10;  B  and 
C,  $12.     How  much  has  each  ? 


SIMPLE  EQUATIONS.  31 

59.  I  have  in  mind  a  certain  number.  If  this  number  is 
diminished  by  8  and  the  remainder  multiplied  by  8,  the 
result  is  the  same  as  if  the  number  were  diminished  by  6 
and  the  remainder  multiplied  by  6.     What  is  the  number  ? 

60.  A  man  having  only  ten-cent  pieces  and  five-cent 
pieces  wished  to  give  some  children  15  cents  each,  but 
found  that  he  had  not  money  enough  by  25  cents;  he, 
therefore,  gave  them  10  cents  each  and  had  30  cents  left. 
How  many  children  were  there  ? 

61.*  A  sum  of  money  was  divided  among  A,  B,  and  C  in 
such  a  way  that  A  received  three  times  as  much  as  B,  and 
B  twice  as  much  as  C.  If  A  received  $6  more  than  C,  how 
much  did  each  receive  ? 

62.  The  sum  of  the  ages  of  a  man  and  his  son  is  80 
years  ;  and  the  father's  age  is  2  years  more  than  twice  the 
age  of  his  son.     What  is  the  age  of  each  ? 

63.  Two  casks  contain  equal  quantities  of  vinegar.  From 
one  cask  37  gallons  are  drawn,  and  from  the  other  7  gallons 
are  drawn.  The  quantity  now  remaining  in  one  cask  is  7 
times  that  remaining  in  the  other.  How  much  did  each 
cask  contain  at  first  ? 

64.  A  merchant  has  two  kinds  of  tea;  one  worth  50 
cents  a  pound,  and  the  other  75  cents  a  pound.  He  makes 
a  mixture  from  these  of  100  pounds,  worth  60  cenfs  a 
pound.     How  many  pounds  of  each  kind  does  he  take  ? 

65.  A  had  $7  and  B  had  $5.  B  gave  A  a  certain  sum ; 
then  A  had  3  times  as  much  as  B.  How  many  dollars  did 
B  give  A  ? 

66.  A  boy  bought  9  dozen  oranges  for  $2.50.  For  a  part 
he  paid  25  cents  a  dozen,  and  for  the  remainder  30  cents  a 
dozen.     How  many  dozen  of  each  kind  did  he  buy  ? 


32  SIMPLE  EQUATIONS. 

Note.    In  the  following  examples  express  in  cents  all  money  values. 

67.  How  can  $2.25  be  paid  in  quarters  and  ten-cent 
pieces  so  as  to  pay  twice  as  many  ten-cent  pieces  as 
quarters  ?  , 

68.  I  have  $1.80  in  ten-cent  pieces  and  five-cent  pieces, 
and  have  four  times  as  many  five-cent  pieces  as  ten-cent 
pieces.     How  many  have  I  of  each  ? 

69.  I  have  $6  in  silver  half-dollars  and  quarters,  and  I 
have  20  coins  in  all.     How  many  have  I  of  each  ? 

70.  I  have  five  times  as  many  half-dollars  as  quarters, 
and  the  half-dollars  and  quarters  amount  to  $11.  How 
many  have  I  of  each  ? 

71.  A  man  has  $65  in  ten-dollar  bills  and  one-dollar 
bills.  He  has  three  times  as  many  one-dollar  bills  as  ten- 
dollar  bills.     How  many  bills  has  he  of  each  kind  ? 

72.  A  sum  of  money  is  divided  among  three  persons, 
A,  B,  and  C,  in  such  a  way  that  A  and  B  together  have 
$6,  A  and  C  $6.50,  and  B  and  C  $7.50.  How  much  has 
each  ? 

73.  A  purse  contains  27  coins  which  amount  to  $11.25. 
There  is  a  certain  number  of  silver  dollars,  and  three  times 
as  many  half-dollars  as  dollars;  the  remaining  coins  are 
quarters.     Find  the  number  of  each. 

74.  A  man  bought  10  yards  of  calico  and  20  yards  of 
cloth  for  $30.60.  The  cloth  cost  as  many  quarters  per 
yard  as  the  calico  cost  cents  per  yard.  Find  the  price  of 
each  per  yard. 

75.  A  man  has  a  certain  number  of  dollars,  half-dollars, 
and  quarters.  The  number  of  quarters  is  twice  the  number 
of  half-dollars  and  four  times  the  number  of  dollars.  If 
he  has  $15,  how  many  coins  of  each  kind  has  he  ? 


CHAPTER  IH. 
POSITIVE  AND  NEGATIVE   NUMBERS. 

61.  Positive  and  Negative  Quantities.  If  a  person  is  en- 
gaged in  trade,  his  capital  will  be  increased  by  his  gains, 
and  diminished  by  his  losses. 

Increase  in  temperature  is  measured  by  the  number  of 
degrees  the  mercury  rises  in  a  thermometer,  and  decrease 
in  temperature  by  the  number  of  degrees  the  mercury  falls. 

In  considering  any  quantity  whatever,  a  quantity  that 
increases  the  quantity  under  consideration  is  called  a  posi- 
tive quantity ;  and  a  quantity  that  decreases  the  quantity 
under  consideration  is  called  a  negative  quantity. 

62.  The  Natural  Series  of  Numbers.  If  from  a  given  point, 
marked  0,  we  draw  a  straight  line  to  the  right,  and  begin- 
ning from  the  zero  point  lay  off  units  of  length  on  this  line, 
the  successive  repetitions  of  the  unit  will  be  expressed  by 
the  natural  series  of  numbers,  1,  2,  3,  4,  etc.     Thus : 


1 

L_ 

2 

1 

3 

1 

4 

1 

5 

1 

6 

1 

7 
1 

8 

1 

9     10    11 

1       1       1 

In  this  series  if  we  wish  to  add  2  to  5,  we  begin  at  5, 
count  2  units  forwards,  and  arrive  at  7.  If  we  wish  to 
subtract  2  from  5,  we  begin  at  5,  count  2  units  backwards, 
and  arrive  at  3.  If  we  wish  to  subtract  5  from  5,  we  count 
5  units  backwards  from  5,  and  arrive  at  0.  If  we  wish  to 
subtract  5  from  2,  we  cannot  do  it,  because  when  we  have 
counted  backwards  from  2  as  far  as  0,  the  natural  series  of 
numbers  comes  to  an  end. 


34  POSITIVE  AND  NEGATIVE  NUMBERS. 

63.  Positive  and  Negative  Numbers.  In  order  to  subtract 
a  greater  number  from  a  smaller  it  is  necessary  to  assume 
a  new  series  of  numbers,  beginning  at  zero  and.  extending 
to  the  left  of  zero.  The  series  to  the  left  of  zero  must 
proceed  from  zero  by  the  repetitions  of  the  unit,  precisely 
like  the  natural  series  to  the  right  of  zero ;  and  the  oppo- 
sition between  the  right-hand  series  and  the  left-hand  series 
must  be  clearly  marked.  This  opposition  is  indicated  by 
calling  every  number  in  the  right-hand  series  a  positive 
number,  and  prefixing  to  it,  when  written,  the  sign  -f- ;  and 
by  calling  every  number  in  the  left-hand  series  a  negative 
number,  and  prefixing  to  it  the  sign  — .  The  two  series  of 
numbers  may  be  called  the  algebraic  series  of  numbers,  and 
written  thus : 


-4      -3-2-1       *0      +1      +  2     -f3      +4 

I I I I I _J I J I 

If,  in  this  double  series  of  numbers,  we  wish  to  subtract 
4  from  2,  we  begin  at  2  in  the  positive  series,  count  4  units 
in  the  negative  direction  (to  the  left),  and  arrive  at  —  2  in 
the  negative  series ;  that  is,  2  —  4  =  —  2. 

The  result  obtained  by  subtracting  a  greater  number  from 
a  less,  when  both  are  positive,  is  always  a  negative  number. 

In  general,  if  a  and  b  represent  any  two  numbers  of  the 
positive  series,  the  expression  a  —  b  will  be  a  positive  num- 
ber when  a  is  greater  than  b ;  will  be  zero  when  a  is  equal 
to  b ;  will  be  a  negative  number  when  a  is  less  than  b. 

In  counting  from  left  to  right  in  the  algebraic  series, 
numbers  increase  in  magnitude ;  in  counting  from  right  to 
left,  numbers  decrease  in  magnitude.  Thus,  —3,-1,  0, 
-f  2,  +  4,  are  arranged  in  ascending  order  of  magnitude. 

64.  The  Absolute  Value  of  a  Number.  The  absolute  value 
of  a  number  is  its  value  independent  of  its  sign. 


POSITIVE  AND  NEGATIVE  NUMBERS.  35 

65.  Every  algebraic  number,  as  +  4  or  —  4,  consists  of 
a  sign  +  or  —  and  the  absolute  value  of  the  number.  The 
sign  shows  whether  the  number  belongs  to  the  positive  or 
negative  series  of  numbers ;  the  absolute  value  shows  the 
place  the  number  has  in  the  positive  or  negative  series. 

When  no  sign  stands  before  a  number,  the  sign  +  is 
always  understood.     But  the  sign  —  is  never  omitted. 

66.  Two  algebraic  numbers  that  have,  one  the  sign  -f-, 
and  the  other  the  sign  — ,  are  said  to  have  unlike  signs.' 

Two  algebraic  numbers  that  have  the   same  absolute 
values,  but  unlike  signs,  cancel  each  other  when  combined. 
Thus, +  4-4  =  0;  +  a  -  a  =  0. 

67.  Double  Meanings  of  the  Signs  +  and  — .     The  use  of 

the  signs  +  and  —  to  indicate  addition  and  subtraction 
must  be  carefully  distinguished  from  the  use  of  the  signs 
+  and  —  to  indicate  in  which  series,  the  positive  or  the 
negative,  a  given  number  belongs.  In  the  first  sense  they 
are  signs  of  operations,  and  are  common  to  Arithmetic  and 
Algebra ;  in  the  second  sense  they  are  signs  of  opposition, 
and  are  employed  in  Algebra  alone. 

Note.  In  Arithmetic,  if  the  things  counted  are  whole  units,  the 
numbers  that  count  them  are  called  whole  numbers,  integral  numbers, 
or  integers,  the  adjective  being  transferred  from  the  things  counted 
to  the  numbers  that  count  them.  But  if  the  things  counted  are  only 
parts  of  units,  the  numbers  that  count  them  are  called  fractional  num- 
bers, or  simply  fractions,  the  adjective  being  transferred  from  the 
things  counted  to  the  numbers  that  count  them. 

Likewise  in  Algebra,  if  the  units  counted  are  negative,  the  numbers 
that  count  them  are  called  negative  numbers,  the  adjective  that  defines 
the  nature  of  the  units  counted  being  transferred  to  the  numbers  that 
count  them. 

68.  Addition  of  Algebraic  Numbers.  An  algebraic  number 
is  often  enclosed  in  a  parenthesis,  in  order  that  the  signs  + 


36  POSITIVE  AND  NEGATIVE  NUMBERS. 

and  — ,  which  are  used  to  distinguish  positive  and  negative 
numbers,  may  not  be  confounded  with  the  +  and  —  signs 
that  denote  the  operations  of  addition  and  subtraction. 

Thus,  +  4  +  (—  3)  expresses  the  sum,  and  +  4  —  (—  3)  expresses 
the  difference,  of  the  numbers  +  4  and  —  3. 

69.  In  order  to  add  two  algebraic  numbers,  we  begin  at 
the  place  in  the  series  which  the  first  number  occupies,  and 
count,  in  the  direction  indicated  by  the  sign  of  the  second 
number,  as  many  units  as  there  are  in  the  absolute  value  of 
the  second  number. 


-4      -3      -2      -1         0      +1      +2     +3      +4 

I I I I I I I t  [ 

Thus,  the  sum  of+2  +  (+3)is  found  by  counting  from 
+  2  three  units  in  the  positive  direction;  that  is,  to  the  right 
and  is,  therefore,  -j-  5. 

The  sum  of  +  2  +  (  —  3)  is  found  by  counting  from  +  2 
three  units  in  the  negative  direction  ;  that  is,  to  the  left,  and 
is,  therefore,  —  1. 

The  sum  of  —  2  -f-  (+  3)  is  found  by  counting  from  —  2 
three  units  in  the  positive  direction,  and  is,  therefore,  +  1. 

The  sum  of  —  2  +  (—  3)  is  found  by  counting  from  —  2 
three  units  in  the  negative  direction,  and  is,  therefore,  —  5. 

70.  If  a  and  b  represent  any  two  numbers,  we  have 
+  a  +  (+  b)  =  a  +  b.  -a  +  (+b)  =  -a  +  b. 
+  a  +  (-b)  =  a-b.         -a  +  (-b)  =  -a-b. 

Therefore,  from  these  four  cases,  we  have  the  following 
Rule  for  Adding  Two  Algebraic  Numbers: 

1.  If  the  numbers  have  like  signs,  find  the  sum  of  their 
absolute  values,  and  prefix  the  common  sign  to  the  result. 

2.  If  the  numbers  have  unlike  signs,  find  the  difference  of 
their  absolute  values,  and  prefix  the  sign  of  the  greater  num- 
ber to  the  result. 


POSITIVE  AND  NEGATIVE  NUMBERS.  37 

71.  The  result  is  called  the  algebraic  sum  in  distinction 
from  the  arithmetical  sum  ;  that  is,  the  sum  of  the  absolute 
values  of  the  numbers. 

Note.  If  there  are  more  than  two  numbers  to  be  added,  add  two 
of  the  numbers,  and  then  this  sum  to  a  third  number,  and  so  on ;  or 
find  the  sum  of  the  positive  numbers  and  the  sum  of  the  negative 
numbers,  then  the  difference  between  the  absolute  values  of  these  two 
sums,  and  prefix  the  sign  of  the  greater  sum  to  the  result. 


Exercise  9. 
Perform  mentally  the  indicated  additions : 
1.  2.  3.  4.  5.  6. 


+  8 

-8 

-7 

-11 

+  11 

-11 

-7 

+  7 

-7 

+  4 

-  4 

-  4 

7. 

8. 

9. 

10. 

11. 

12. 

+  5 

-8 

-9 

-20 

+  87 

-37 

-7 

-7 

-7 

+  24 

-36 

+40 

-9 

+  9 

-6 

+  36 

-42 

-20 

13. 

14. 

15. 

16. 

17. 

18. 

+  15 

-15 

-21 

-20 

-18 

+  17 

+  12 

-12 

+  12 

-30 

+  32 

-27 

-20 

+  30 

-13 

+  40 

-12 

-19 

19. 

20. 

21. 

22. 

23. 

24. 

31 

-31 

81 

19 

-15 

90 

-17 

-17 

-71 

-  7 

-70 

-30 

-15 

-15 

31 

13 

-40 

-20 

21 

21 

-40 

-  28 

-90 

10 

38  POSITIVE  AND  NEGATIVE  NUMBERS. 

Addition  of  Similar  Monomials. 

1.  Find  the  sum  of  3  a,  2  a,  a,  5  a,  7  a. 

The  sum  of  the  coefficients  is  3  +  2  +  1+5  +  7  =  1 8. 
Hence,  the  sum  of  the  monomials  is  18  a. 

2.  Find  the  sum  of  —  5  c,  —  c,  —  3  c,  —  4  c,  —2  c. 

The  sum  of  the  coefficients  is  —  5  —  1—  3  —  4  —  2  =  —  15. 
Hence,  the  sum  of  the  monomials  is  —  15  c. 

3.  Find  the  sum  of  Sx,  —  9x,  —  x,  3x,  &x,  —  12  x,  x. 

The  sum  of  the  positive  coefficients  is8  +  3  +  4  +  l  =  16. 
The  sum  of  the  negative  coefficients  is  —  9  —  1  —  12  =  —  22. 
The  difference  between  16  and  22  is  6,  and  the  sign  of  the  greater 
is  negative.     Hence,  the  sum  is  —  6  x.     Therefore, 

72.    To  Find  the  Sum  of  Similar  Monomials, 

Find  the  algebraic  sum  of  the  coefficients,  and  annex  to 
this  sum  the  letters  common  to  the  terms. 

Exercise  10. 
Perform  mentally  the  indicated  additions : 


1. 

2. 

3. 

4. 

5. 

6. 

la 

7xy 

3yz 

2ab 

13  c 

3xz 

2a 

2xy 

-9yz 

-9ab 

12  c 

7xz 

—  5a 

-3xij 

-2yz 

-7ab 

-24  c 

—  9xz 

—  2a 

±xy 

Tyz 

Sab 

15  c 

—     xz 

7. 

8. 

9. 

10. 

11. 

12. 

12  «4 

4  m2 

11  abc 

27  yV 

17  x9 

18/ 

-    7*4 

-9m2 

3abc 

-  41  y*z* 

-31a;8 

27  y* 

-    8z* 

-7m2 

—    7 abc 

-   2yV 

-47*8 

-43/ 

-       z* 

5  m2 

—    9  abc 

-     y4*4 

61  z8 

-  21  ?/ 

POSITIVE  AND  NEGATIVE  NUMBERS.  39 

Exercise  11. 
Eind  the  algebraic  sum  of : 

1.  7  a,  2  a,  —  3  a,  —  5  a.  9.  3  yz,  —  9  yz,  20  yz,  7  ys. 

2.  7  ay,  2  ccy,  —  4  #y,  —  5  xy.  10.  2  a£,  —  10  ab,  —  lab,3  ah. 

3.  4  a2£,  -  3  a26,  -  5  a2b.         11.  14  z2,  9  x2,  -  8  x2,  -  11  x\ 

4.  3xy,4xy,7xy,-3xy.      12.  17  ys,  -  8 y8,  5  y\  -  f. 

5.  16&,  -11&,  -25,  3&.        13.  12*4,  -7z4,  -Sz4,  -9s4. 

6.  13c,  12c,  -24c,  2c.  14.  2\y2,-\7y2,-3f,-^f. 

7.  —  3fc*,7a»,  —  2a»,  —  a*.  15.  c2,  -  2  c2,  -  15  c2,  - 18  cfl. 

8.  2  ac,  5  ac,  —  9  ac,  3  ac.        16.  4m2,  —  11  m2,  —  7m2, 5m2. 

Express  in  one  term  each  of  the  following : 

17.  9z2-7a:2  +  4a:2-3a2  +  3a;2-5a;2. 

18.  3  a2  -  18  a2  +  a2  -  5  a2  +  6  a2  -  10  a2. 

19.  5  A  +  7  a8z  -  9  a*x  -  29  a*u  +  4  a8x. 

20.  -  5  a262  +  7  a2b2  +  11  a2*2  -  4  a2*2  -  9  a2£-. 

21.  -  21  axs  +  20  ar5  -  6  aa8  4-  5  az8  -  13  ax3. 

22.  —  11  abcx  +  3  a£c#  —  7  a£ca;  +  29  a6c#  +  obex. 

23.  -  3  y4z4  -  27  yV  -  2  yV  +  41  #V  4  y4z4. 

24.  -  4  xY  +  18  afy5  4-  27  x6y*  -  43  aV  -  x6if. 

25.  -  31  a&28  4- 17  abz*  -  47  abz3  4  61  a^8  +  abz\ 

26.  ia  +  |a-T^  +  |a-fa  +  a. 

27.  |a2-5a2-f  |a2  +  7a2-f  a24-£a2. 

28.  xy-\xy  +  \xy-\xy  +  \xy  —  \\xy. 

29.  }«_|s_f«4-3  +  }.|s_  js. 

30.  ^  ayz  -\-  ty-  ays  4-  ty  ayz  +  *■£■  ayz  —  7  ays. 

31.  14  a  -  la  -  3a-  15a  +  12a. 

32.  4  a2c  -  10  a2c  +  6  a2c  -  9  a2c  4-  a2c. 
S3.  3aty-4afy  +  2afy-a;V  +  5a;y 


40  POSITIVE  AND  NEGATIVE  NUMBERS. 


Subtraction  of  Algebraic  Numbers. 

73.  In  order  to  subtract  one  algebraic  number  from 
another,  we  begin  at  the  place  in  the  series  which  the 
minuend  occupies,  and  count,  in  the  direction  opposite  to 
that  indicated  by  the  sign  of  the  subtrahend,  as  many  units 
as  there  are  in  the  absolute  value  of  the  subtrahend. 


-4     -3     -2     -1         0     +1      +  2     +3     +4 

I I I ! ! I I I I 

Thus,  the  result  of  subtracting  +  1  from  +  3  is  found 
by  counting  from  +  3  one  unit  in  the  negative  direction  ; 
that  is,  in  the  direction  opposite  to  that  indicated  by  the  sign 
+  before  1,  and  is,  therefore,  +  2. 

The  result  of  subtracting  —  1  from  +  3  is  found  by 
counting  from  +  3  one  unit  in  the  positive  direction,  and 
is,  therefore,  -f  4. 

The  result  of  subtracting  4-  1  from  —  3  is  found  by 
counting  from  —  3  one  unit  in  the  negative  direction,  and 
is,  therefore,  —  4. 

The  result  of  subtracting  —  1  from  —  3  is  found  by 
counting  from  —  3  one  unit  in  the  positive  direction,  and 
is,  therefore,  —  2. 

If  a  and  b  represent  any  two  numbers,  we  have 
+  a  —  (+b)  =  a-b.  -  a  -  (+b)  =  -  a -b. 

+  a-\-b)  =  a  +  b.  -a-\-b)  =  -a  +  b. 

74.  From  these  four  eases  we  see  that  subtracting  a  posi- 
tive number  is  equivalent  to  adding  an  equal  negative  num- 
ber ;  and  that  subtracting  a  negative  number  is  equivalent 
to  adding  an  equal  positive  number.     Therefore, 

75.  To  Subtract  One  Algebraic  Number  from  Another, 
Change  the  sign  of  the  subtrahend,  and  add  the  result  td 

the  minuend. 


POSITIVE  AND  NEGATIVE  NUMBERS.  41 

Exercise  12. 
Perform  mentally  the  indicated  subtractions : 


1. 

2. 

3. 

4. 

5. 

6. 

11 

-11 

-11 

11 

3 

-    3 

3 

-    3 

3 

-   3 

11 

-11 

7. 

8. 

9. 

10. 

11. 

12. 

3 

3 

-3 

3 

-3 

3 

8 

-8 

-8 

-9 

-9 

9 

13. 

14. 

15. 

16. 

17. 

18. 

-8 

-7 

8 

7 

7 

6 

7 

8 

-7 

8 

-3 

-4 

Subtraction  of  Similar  Monomials. 

1.   From  15m2x2  take  -  7  m2x2. 

15  mW  -  (-  7  m2^2)  =  15  m2x2  +  7  m^x2 
=  22  m2x2.     Hence, 

76.    To  Subtract  a  Monomial  from  a  Similar  Monomial, 

Change  the  sign  of  the  coefficient  of  the  subtrahend ;  then 
add  the  coefficients,  and  annex  the  common  letters  to  the  result. 


, 

Exercise  13. 

Perform  mentally  the  indicated  subtractions : 

1.                        2. 

3. 

4. 

5. 

165-^vwvw,      3yz 
-115<^,-9y2 

- 10  ab 

-    6  a2 

-7m2 

-    lab 

-10  a2 

3  m2 

6.                      7. 

8. 

9. 

10. 

-29  c              -f41c 

-29c 

Sabe 

—  ac2x 

—  41c                  29c 

-41c 

—  abc 

Sac2x 

42  POSITIVE  AND  NEGATIVE  NUMBERS. 

If  a  =  4,  6  =  —  2,  c  =  —  3,  find  the  value  of: 

11.  a  +  (-  b)  +  c.  14.    -  (-  a)  +  (-  b)  -  (-  c). 

12.  -a-(-b)  +  c.  15.    +  (-a)-(-&)-(-c). 

13.  a-b  +  (-c).  16.    -(-«)-(-£)  +  (-/). 

Multiplication  of  Algebraic  Numbers. 

77.  Multiplication  is  generally  defined  in  Arithmetic  as 
the  process  of  finding  the  result  when  one  number  (the 
multiplicand)  is  taken  as  many  times  as  there  are  units  in 
another  number  (the  multiplier).  This  definition  fails 
when  the  multiplier  is  a  fraction.  In  multiplying  by  a 
fraction,  we  divide  the  multiplicand  into  as  many  equal 
parts  as  there  are  units  in  the  denominator,  and  take  as 
many  of  these  parts  as  there  are  units  in  the  numerator. 

If,  for  example,  we  multiply  6  by  § ,  we  divide  6  into 
three  equal  parts  and  take  two  of  these  parts,  obtaining  4 
for  the  product.  The  multiplier,  f ,  is  §  of  1,  and  the  prod- 
uct, 4,  is  £  of  6 ;  that  is,  the  product  is  obtained  from  the 
multiplicand  precisely  as  the  multiplier  is  obtained  from  1. 

78.  Multiplication  may  be  defined,  therefore, 

As  the  process  of  obtaining  the  product  from  the  multipli- 
cand as  the  multiplier  is  obtained  from  unity. 

79.  Every  extension  of  the  meaning  of  a  term  must 
be  consistent  with  the  sense  previously  attached  to  the 
term,  and  with  the  general  laws  of  numbers. 

This  extension  of  the  meaning  of  multiplication  is  con- 
sistent with  the  sense  attached  to  multiplication  when  the 
multiplier  is  a  positive  whole  number. 

Thus,  5X7  =  35, 

the  multiplier,  5,  =1  +  1  +  1  +  1-1-1, 

and  the  product,  35,  =7  +  7  +  7  +  7  +  7. 


POSITIVE  AND  NEGATIVE  NUMBERS.  43 

Law  of  Signs  in  Multiplication. 
By  the  definition  of  multiplication  (§  78), 
since  +  3  =  +  l  +  l  +  l, 

3  x  (+  8)  =  +  8  +  8  +  8  =  4-  24, 
and  3  X  (-  8)  =  -  8  +  (-  8)  +  (-  8)  =  -  24. 

Again,  since       — 3  =  —  1  —  1  —  1, 

(_  3)  X  8  =  -  8  -  8  -  8  =  -  24, 
and        (-  3)  X  (-  8)  =  -  (-  8)  -  (-  8)  -  (-  8) 
=  +  8 +  8 +  8= +  24. 

The  minus  sign  before  the  multiplier,  3,  signifies  that  the 
repetitions  of  the  multiplicand  are  to  be  subtracted. 
If  a  and  b  stand  for  any  two  numbers,  we  have 
(+  a)  X  (+  b)  =  +  ab, 
(H-  a)  X  (—  b)  =  —  ab, 
(—  a)  X  (+  b)  =  —  ab, 
,     (-  a)  X  (-  b)  =  +  ab. 
That  is,  if  two  numbers  have  like  signs,  the  product  has  the 
plus  sign ;  if  unlike  signs,  the  product  has  the  minus  sign. 

80.  The  Law  of  Signs  in  Multiplication  is,  therefore, 
Like  signs  give  +,  and  unlike  signs  give  —. 

The  Index  Law  in  Multiplication. 

Since  a2  =  aa,  and  a%  =  aaa, 

a2  X  az  =  aa  X  aaa  =  ab  =  a2+i; 
a4  X  a  =  aaaa  X  a  =  ah  =  a4+1. 

If  then  m  and  n  are  positive  integers, 
am  X  an  =  am  +  n. 

In  like  manner,  am  X  aM  X  ap  =  aw  +  ra+*>. 

81.  The  Index  Law  in  Multiplication  is,  therefore, 

The  exponent  of  a  letter  in  the  product  is  equal  to  the  sum 
of  the  exponents  of  the  letter  in  the  factors  of  the  product. 


44  POSITIVE  AND  NEGATIVE  NUMBERS. 

Multiplication  of  Monomials. 

1.  Find  the  product  of  6a2b*  and  7  ab2c*. 

Since  the  order  of  the  factors  is  immaterial,  (§  42) 

6  a268  x  7  abW  =  6x7x  a2  xaxbzxWxc* 
=  42  aWc*. 

2.  Find  the  product  of  —  3  ab  and  7  ab8. 

—  Sab  X7  abz  =  — 3x7  XaXaXbxb*  ** 

=  -  21  a26*. 

3.  Find  the  product  of  #n  and  x8,  and  of  ccn  and  xn. 

x»  x  x8  —  xn  +3. 
xn  X  xn  =  xn+n  =  x2n.     Therefore, 

82.  To  Find  the  Product  of  Two  Monomials, 

Find  the  product  of  the  numerical  coefficients ;  and  to 
this  product  annex  the  letters,  giving  to  each  letter  an  ex- 
ponent equal  to  the  sum  of  its  exponents  in  the  factors. 

83.  A  product  of  three  or  more  factors  is  called  the  con- 
tinued product  of  the  factors. 

1.  Find  the  continued  product  of  (—  a)x(—  b)X(—  c). 

By  the  law  of  signs,  §  80,  we  have 

(-  a)  x  (-  b)  =  ab, 
and  (ab)  X  (—  c)  =  —  abc. 

C 

2.  Find  the  continued  product  of 

(-a)  X(-b)X  (-c)  X  (-«*)• 
By  the  law  of  signs,     (—a)  x  (—  6)  =  ab, 

(ab)  X  (—  c)  =  —  abc, 
(—  abc)  x  (—  d)  =  abed. 

84.  From  Examples  1  and  2  (§  83),  we  see  that  an  odd 
number  of  negative  factors  gives  a  negative  product;  and  an 
even  number  of  negative  factors  gives  a  positive  product. 


POSITIVE  AND  NEGATIVE  NUMBERS.  45 


Exercise  14. 

Note.  The  beginner  should  first  write  the  sign  of  the  product; 
then  the  product  of  the  numerical  coefficients  after  the  sign ;  and, 
lastly,  the  letters  in  alphabetical  order,  giving  to  each  letter  the 
proper  exponent. 


Find  mentally  the  product  of : 


1. 

2. 

3. 

4. 

5. 

6. 

3a 

-3a 

3a 

~3a 

9  a2 

-9a2 

2a 

-2a 

-2a 

2a 

6a 

6  a2 

7. 

8. 

9. 

10. 

11. 

12. 

5x2 

-7  a* 

8  a8 

-7m2 

x2y2 

-  a  W 

lxh 

—     a 

-7  a8 

8  m4 

—  5  x2y2 

a2bc  - 

dr^j-W** 

13. 

14. 

15. 

16. 

17. 

18. 

-3a2x 

5<zc3 

7x5 

-9b2 

-a*b 

9am 

-3  ax2 

6a2c 

-3z7 

-6a2 
2    &V* 

-ah* 

6a» 

19. 

20. 

21. 

22. 

23. 

24. 

am+1 

—  xn 

zn+s 

a" 

2/" 

<xm_1 

2xm~ 

zn-2 

x2y,  xhf, 

a                 yn+2 
and   —  15  abxy.  ^  3.0  It 

-q^'T*3' 

-  z-x 

25.    -2 

26.    -8 

a2,  -2  b2, 

—  3  ab,  and  a2b2. 

27.  ab,  —  ac,  —  be,  and  —  3  abe. 

28.  —  xyz2,  x2yz,   —  xy2z,  and  3  xyz. 

29.  4  a3,  -10a26,  25  a&2,  and  -  ab\ 

30.  62,  6a&2,  -±a2b,  and  -2a&2. 

31.  a8n,  a2""1,  a2*+1,  a1"2",  and  a. 

32.  xm+1,  xm-'z,  xm,  xm+n,  and  xm~n. 


46  POSITIVE  AND  NEGATIVE  NUMBERS. 


Division  of  Algebraic  Numbers. 

85.  Division  is  the  operation  of  finding  one  of  two 
factors,  when  their  product  and  the  other  factor  are  given. 

86.  With  reference  to  this  operation  the  product  is  called 
the  dividend,  the  given  factor  the  divisor,  and  the  required 
factor  the  quotient. 

Law  of  Signs  in  Division. 
Since  (+  a)  X  (+  b)  =  +  ab,  .*.  +  ab  -r-  (+  a)  =  +  b. 
Since  (+  a)  X  (—  b)  =  —  ab,  .'.  —  ab  -i-  (+  a)  =  —  b. 
Since  (—  a)  X  (-f-  b)  =  —  ab,  .*.  —  ab  •*-  (—  a)  —  +  b. 
Since  (—  a)  X  (—  b)  =  +  ab,  .'.  +  ab  -5-  (—  a)  =  —  b. 

That  is,  if  the  dividend  and  divisor  have  like  signs,  the 
quotient  has  the  plus  sign ;  and  if  they  have  unlike  signs, 
the  quotient  has  the  minus  sign. 

87.  The  Law  of  Signs  in  Division  is,  therefore, 
Like  signs  give  +,  and  unlike  signs  give  — . 

The  Index  Law  in  Division. 

The  quotient  contains  the  factors  of  the  dividend  that 
are  not  found  in  the  divisor. 

1.    Divide  a5  by  a2.  2.    Divide  aA  by  a. 

•     a6       aaaaa  ft         .    „ 

1.    —  = =  aaa  =  a9  =  ab~  . 

a*  aa 


a*      aaaa 


J- 

2.    —  = =  aaa 

a  a 


If  m  and  n  are  positive  integers,  and  m  is  greater  than  n, 
am  -T-  an  =  am~n. 


POSITIVE  AND  NEGATIVE  NUMBEBS.  47 

88.    The  Index  Law  in  Division  is,  therefore, 

The  exponent  of  a  letter  in  the  quotient  is  equal  to  the 

exponent  of  the  letter  in  the  dividend  minus  the  exponent  of 

the  letter  in  the  divisor. 

Division  of  Monomials. 

1.  Divide  24  a7  by  8  a6. 

8  a6 
We  obtain  the  factor  3  of  the  quotient  by  dividing  24  by  8  ;  and  the 
factor  a2  of  the  quotient,  by  writing  a  with  an  exponent  equal  to  the 
exponent  of  a  in  the  dividend  minus  the  exponent  of  a  in  the  divisor. 

2.  Divide  20  a5b6  by  -  4  ab\ 

20a»P  _    ,    ...    ,  ... 

— 7—7;  =  —  5a5~166-6  =  —  5a4&. 

3.  Divide  -30aW  by  ~20abc\ 

-30aWe* 

f  aWc. 


—  20  abc* 

4.   Divide  -  57 a*"1  by  —  19  a*-3. 

— -Lz — -  =  3^-1-^-3)  =  3a*-i-*+3  =  3aa# 
—  19  ax~ 3 

5!   Divide  77  a?mbn<f-  by  11  am£Bc8. 

77  a2"1^* 


11  fl^C3 


=  7  a2m-m  &»— n  c*-3  =  7  o^c*— 8. 


6n  l>n 

Note.     Since  by  division  —  =  1 ;  and  by  the  index  law  —  =  &°,  it 

follows  that  6°  as  1.  Hence,  any  letter  which  by  the  rule  would  appear 
in  the  quotient  with  zero  for  an  exponent,  may  be  omitted  without  affect- 
ing the  quotient. 

89.    To  Find  the  Quotient  of  Two  Monomials,  therefore, 
Divide  the  numerical  coefficient  of  the  dividend  by  the 
numerical  coefficient  of  the  divisor  ;  and  to  the  result  annex 
the  letters,  giving  to  each  letter  an  exponent  equal  to  its 
exponent  in  the  dividend  minus  its  exponent  in  the  divisor. 


48  POSITIVE  AND  NEGATIVE  NUMBEBS. 

Exercise  15. 
Perform  mentally  the  indicated  divisions : 


1. 

2. 

3. 

4. 

5. 

12  ab*    xi* 
6ab 

16  aV     . 
8ae 

20  ay  ^.x 
oxy       u 

14  a?y  £8 
7  ay«8 

54  a*ic8c4 
6  a8ic2c 

6. 

7. 

8. 

9. 

10. 

-  20  cy 

-  27  a4z2 

—  56  aAxA 
-8a*x 

63  ^z8 

-7^2 

72  x*yz2 

5c2y 

-9  a3 

12  xy« 

11. 

12. 

13. 

14. 

15. 

-  9  a2b  V 

-  27  ab2xs 
Sabx2 

-  3  xY zK 

-2cy 

-18xn 

—  ab2c 

x2y2zz 

-4a;"-s 

16. 

17. 

18. 

19. 

20. 

56  a2xY 
-lx*tf 

-  3  ab  V 
-  a6V 

-  2  a262z' 
ate2 

1           4aV 

-2«y 

6aV 
-2aar* 

21. 

22. 

23. 

24. 

25. 

-a5x9 
-a*x* 

26. 

-51aV 
17  ay 

27. 

28  c»d4 
-Itfd2 

28. 

16  afc2*3 
—  4  a#2a: 

29. 

12  icVV 
-  3'xy2z* 

30. 

7  a2fon 
7  afo"-1 

3an  +  1         : 

L2  £cni/n         54 
6x2y 

a""2?/"-1 

-9«y 

xY 

xn~2yn~l 

31. 

32. 

33. 

34. 

35. 

-6cy 

36. 

4  aAbA 
2a2b2 

37. 

-3a4& 
38. 

2y*z 
39. 

—  i/r^8 
40. 

-8Cy 

10  aV 
5aV 

2- 

12  al2lf 
1* 

3ay° 
—  xY 

-  x'V 

-4cy 

t/i1 

CHAPTER  IV. 
ADDITION  AND   SUBTRACTION. 


Integral  Compound  Expressions. 

90.  If  an  algebraic  expression  contains  no  letter  in  the 
denominator  of  any  of  its  terms,  it  is  called  an  integral 
expression.  Thus,  xz  -f-  7  ex2  —  c8  —  5  c2x,  -J-  ax  —  £  bey,  are 
integral  expressions. 

An  integral  expression  may  have  for  some  values  of  the  letters  a 
fractional  value,  and  a  fractional  expression  an  integral  value.  If, 
for  instance,  a  stands  for  £  and  6  for  £,   the  integral  expression 

2  a  — 5  b  stands  f or  £  —  f  =  \ ;  and  the  fractional  expression  —  stands 

for  ^  -r  £  =  5.  Integral  and  fractional  expressions,  therefore,  are  so 
named  on  account  of  the  form  of  the  expressions,  and  with  no  refer- 
ence whatever  to  the  numerical  value  of  the  expressions  when  definite 
numbers  are  put  in  place  of  the  letters. 

Addition  of  Integral  Compound  Expressions. 

91.  The  addition  of  two  compound  algebraic  expressions 
can  be  represented  by  connecting  the  second  expression 
with  the  first  by  the  sign  +.  If  there  are  no  like  terms  in 
the  two  expressions,  the  operation  is  algebraically  complete 
when  the  two  expressions  are  thus  connected  (§  11,  Note). 

If,  for  example,  it  is  required  to  add  m  -+-  n  —  p  to 
a  +  b  +  c,  the  result  will  be  a  -f-  b  +  c  ■+-  (m  +  n  —  p)  ;  or, 
removing  the  parenthesis  (§  39),  a  +  b-\-c  +  m  +  n—  p. 


50  ADDITION  AND  SUBTRACTION. 

92.  If,  however,  there  are  like  terms  in  the  expressions, 
every  set  of  like  terms  can  be  replaced  by  a  single  term 
with  a  coefficient  equal  to  the  algebraic  sum  of  the  coeffi- 
cients of  the  like  terms. 

1.  Add  5  a2 +  ±a  +  3  to  2  a2 -3a- ±. 

2a2-3a-4-f(5a2  +  4a  +  3) 

=  2a2-3a-4:  +  5a2  +  ±a  +  3  (§39) 

=  2a2  +  5a2-3a  +  4a-4  +  3  (§38) 

=  7  a2  +  a  —  1. 

This  process  is  more  conveniently  represented  by  arrang- 
ing the  terms  in  columns,  so  that  like  terms  shall  stand  in 
the  same  column,  as  follows  : 

2  a2  -  3  a  -  4 
5  a2  +  4  a  -f  3 
7  a2  +     a  -  1 

The  coefficient  of  a2  in  the  result  will  be  5  +  2,  or  7  ;  the  coeffi- 
cient of  a  will  be  —  3  +  4,  or  1 ;  the  last  term  will  be  —  4  +  3,  or  —  1. 

2.  Ad&2a3-3a2b  +  ±ab2  +  b3)  a3  +  ±a2b  -  lab2-  2b3; 
3a?  +  a2b-3  ab2  -  4  b3 ;  and  2  as  +  2  a2b  -f-  6  a£2  -  3  b3. 

2as-3a2b  +  4,ab2  +  b3 
a*  +  ±a2b-l  ab2  -2  b3 
3a3  +  a2b-3ab2-4b* 
2a3  +  2a2b  +  6ab2-3b3 
8a3  +  4:a2b  -8  b9 

The  coefficient  of  a8  in  the  result  will  be  2  +  1  +  3  +  2,  or  +  8  ;  the 
coefficient  of  a26  will  be  —  3  +  4  +  1  +  2,  or +4;  the  coefficient  of 
ab2  will  be  4  —  7  —  3  +  6,  or  0,  and,  therefore,  the  term  ab2  will  not 
appear  in  the  result  (§  21) ;  and  the  coefficient  of  6s  will  be  1  —  2  —  4 
—  3,  or  —  8. 


ADDITION  AND  SUBTRACTION.  51 

Exercise  16. 


1.    a  -+-  b  ;    a  —  b. 


Add: 

3.  5z2  +  6z-2;    3z2-7a;4-2.  #*    -^ 

4.  3a;2-2^?/  +  2/2;    a;a  -  2  a^  4- 3  y*.    ^Y^-  ^  X^  *"  "^ 

5.  ax2  +  ^-4;   3az2-2kc44;    -  4az2  -  2fo  4- 5.    -Z'h'T 

6.  5x  +  3y  +  z-,   3x  +  2y  +  3z;   x-3y-5z. 

7.  -  3  a£  -  2  az2  +  3  A  +  x3 ;    -  4  ab  -  6  A  +  5  ax2 ; 

#3  —  ab  -f-  a2#  —  ace2 ;    ax2  +  8ab  —  5  a2#. 

8.  a4-  2  a8  +  3  a2-  a  +  7;    2  a4  -  3  a8  +  2  a2  -  a  +  6; 

a4-2a3  +  2as-5. 

9.  3a2-a0  +  ac-3&2  +  4fo-s2;    -  Abe  +  5c2  +  2  ao; 

5  a2  -  a&  -  ac  +  5  be  ;    -  4  a2  +  M  -  5  be  4 -  2  c2. 

10.  x4-3z3  +  2z2-4;c  +  7;    3x*  +  2x3  4-  a2  -  5a  -  6  ; 

4;z4  +  3z8-3z2  +  9z-2;    2z4  -  Xs  +  a2  -  a  +  1. 

11.  72/3-3V-4^V  +  ^3;  -Sar'-llz?/2-^^2-?/8; 

x*y  —  xz2  —  y2  —  Bxy2 ;  —  &xz2  +  y2—z3  -\-6xy2  +  10  a;3. 

12.  a4  -  2  a3  +  3  a2  -  3  a  -  2  ;    a3  +  a8  +  a  -  3  a*  +  3 ; 

4  a4  +  5  a8 ;    2a2+3a-2;    -a2  -  2  a  -  3. 

13.  x3  +  2xy2  -  x2y  -  y3;    2x3  -  3x2y  -  4^  -  7^; 

x3  -  8  V  -  7  s/8. 

14.  c4  -  3  c3  +  2  c2  -  4  c  +  7 ;    2  c4  +  3  c3  4-  2  c2  +  5  c  4-  6  ; 

c8-4c4-4c2-5. 

15.  3  x2  —  xy  +  #2  —  3  y2  —  £2 ;   —  5  x2  —  ay  —  xz  +  5  2/2  ; 

tf  +  3yz  4-3*2;  6x2  -  6y  -  6  z  +  ±xz-,  ±yz-5xz. 

16.  m*  —  3  m4n  —  6  ?^8ti2  ;    m8w2  +  raV*  —  5  w4w  -  3»5; 

2mH4m¥-3mw4-w5;    -2mV-3mw4  +  n4; 
-  m*  4-  2  m?i4  4-  2  rc5  +  3  m*n. 

17.  6y-l-2a:2y;    5 +  2xy* -  ±x2y;   -afy-5  +  6a;y*i 

2  +  x^-^j   x2i/-2V-5/  +  l. 


52  ADDITION  AND  SUBTRACTION. 

Subtraction  of  Compound  Expressions. 

93.  The  subtraction  of  one  expression  from  another,  if 
none  of  the  terms  are  alike,  can  be  represented  only  by 
connecting  the  subtrahend  with  the  minuend  by  means  of 
the  sign  — . 

If,  for  example,  it  is  required  to  subtract  a  -\-b  +  c  from 
m,  +  n  —  p,  the  result  will  be  represented  by 

m  +  n—  p  —  (a  +  b  -\-  c); 

or,  removing  the  parenthesis  (§  40), 

m  -\-  n  —  p  —  a  —  b  —  c. 

If,  however,  some  of  the  terms  in  the  two  expressions 
are  alike,  we  can  replace  like  terms  by  a  single  term : 

1.  Subtract  a8  -  2a2 -\- 2a -1  from  3 a*  -  2  a2  +  a  -  2 ; 
the  result  may  be  expressed  as  follows  : 

3  a8  -  2  a2  +  a  -  2  -  (a8.-  2  a2  +  2  a  -  1)  ; 
or,  removing  the  parenthesis  (§  40), 

3a8-2a2  +  a-2-a8  +  2a2-2a  +  l 

=  3a8-a3-2a*  +  2a2  +  a-2a-2  +  l     (§38) 
=  2**-a-l. 

This  process  is  more  easily  performed  by  writing  the 
subtrahend  below  the  minuend,  mentally  changing  the  sign 
of  each  term  in  the  subtrahend,  and  adding  the  two  expres- 
sions.    Thus,  the  above  example  may  be  written 
3a8-2a2+     a-2~-^> 

as-2a2  +  2a-l^%' 
2  a8  -     a  -  1 

The  coefficient  of  a8  will  be  3  —  1,  or  2  ;  the  coefficient  of  a2  will 
be  —  2  +  2,  or  0,  and  therefore  the  term  containing  a2  will  not  appear 
in  the  result ;  the  coefficient  of  a  will  be  1  —  2,  or  —  1 ;  the  last  term 
will  be  —  2  +  1,  or  —  1. 


ADDITION  AND  SUBTRACTION.  53 

2.    Subtract  a5  4  4  a8a2  -  3  A8  -  4  ax4 

from  a8#2  +  2  a2x3  —  4  owe4. 
Here  terms  that  are  alike  can  be  written  in  columns : 
A2  +  2  a2x3  -  4  aa;4 
a5  +  4  A2  -  3  cV  -  4  as4 
-  a6  -  3  A2  +  5  a  V 

There  is  no  term  a5  in  the  minuend,  hence  the  coefficient  of  a6  in 
the  result  will  be  0  —  1,  or  —  1  ;  the  coefficient  of  asx2  will  be  1  —  4,  or 
—  3 ;  the  coefficient  of  a?xz  will  be  2  +  3,  or  +  5 ;  the  coefficient  of 
ax4  will  be  —  4  +  4,  or  0,  and  ax4  will  not  appear  in  the  result. 

Exercise  17. 

1.  From  8a  —  45  —  2c  take  2a  — 3b  — 3c. 

2.  From  3a-4b  +  3c  take  2a-Sb-c-d. 

3.  From  7  a2  -  9  x  -  1  take  5  a2  -  6  x  -  3. 

4.  From  2  a?2  —  2  ace  +  a2  take  x2  —  ax  —  a2. 

5.  From  4 a  —  3 b  —  3c  take  2a  —  3b  +  ±c. 

6.  From5z24-7;z  +  4  take  3z2  -  7x  4  2. 

7.  From  2ax  +  3by  +  5  take  3  a«  —  3  by  —  5. 

8.  From  4  a2  -  6  a&  +  2  62  take  3a2-\-ab-\-  b\ 

9.  From  4  a26  +  7  a&2  +  9  take  8-3  aft2. 

10.  From  5  a2c  +  6  a2b  -8  a8  take  b3  4  6  a2&  -  5  a2c. 

11.  From  a2  -  b2  take  S2.  13.    From  b2  take  a2  -  b\ 

12.  From  a2  -  b2  take  a2.  14.   From  a2  take  a2  -  b2. 
15.  From  a4  +  3  ax3  -  2  foe2  +  3  ex  -  4  d 

take  3  cc4  4-  ax3  —  4:  bx2  +  6  ex  +  d. 

IfA  =  3a2-2ab  +  5b\  C  =  7  a2  -  8  ab  4  5  ft*, 

^  =  9  a2  -  5  oft  4-  3  b2,  D  =  11  a2  -  3  ab- 4b2, 
find  the  expression  for  : 

1Q.   A+C  +  B-D.  19.   A+C-  B-  D. 

11.    A-  C-B  +  D.  20.    A-  C  +  B  +  D. 

18.    C-  A-  B  +  D.  21.   ^  4  C  -  B  4-  2>. 


54  ADDITION  AND  SUBTRACTION 

Insertion  of  Parentheses. 
We  hare  the  following  equivalent  expressions  : 

a  -f-  (b  +  c)  =  a  +  b  +  o,  .'.a  +  Hc  =  fl  +  (H«)l 

a  +  (b  —  c)  =  a  +  ^  —  c,  .*.  a  +  &  —  c  —  a  +  (£  —  c)  ; 

a  —  (b  -f  c)  =  a  —  b  —  c,  .'.  a  —  b  —  c  =  a  —  (b  +  c); 

a  —  (b  —  c)  =  a  —  b  -f-  c,  .'.a  —  b  -{-  c  =  a  —  (b  —  c). 

94.  Hence,  a  parenthesis  preceded  by  the  sign  +  may 
not  only  be  removed  without  changing  the  sign  of  any  term, 
but  may  also  be  inserted,  enclosing  any  number  of  terms, 
without  changing  the  sign  of  any  term. 

And  a  parenthesis  preceded  by  the  sign  —  may  not  only 
be  removed,  provided  the  sign  of  every  term  within  the 
parenthesis  is  changed,  namely,  +  to  —  and  —  to  -+-,  but 
may  also  be  inserted,  enclosing  any  number  of  terms,  pro- 
vided the  sign  of  every  term  enclosed  is  changed. 

95.  Expressions  may  occur  having  parentheses  within 
parentheses.  In  such  cases  signs  of  aggregation  of  dif- 
ferent shapes  are  used,  and  the  beginner,  when  he  meets 
with  one  branch  of  a  parenthesis  (,  or  bracket  [,  or  brace 
\ ,  must  look  carefully  for  the  other  branch,  whatever  may 
intervene;  and  all  that  is  included  between  the  twc 
branches  must  be  treated  as  the  +  or  —  sign  before  the 
sign  of  aggregation  directs.  It  is  best  to  remove  each  paren- 
thesis in  succession,  beginning  with  the  innermost.     Thus, 

1.      a  -  \b  -  (c  -  d)  +  e] 
=  a-[J-c  +  rf-fe] 
=  a  —  b  +  c  —  d  —  e. 

=  a-\b-[c-d  +  e+f]\ 
=  a—\b  —  c  +  d  —  e—f\ 
=  a  —  b  +  c  —  d  +  e  +/. 


ADDITION  AND  SUBTRACTION. 


55 


Exercise  18. 

Simplify  the  following  by  removing  the  parentheses  and 
combining  like  terms : 

1.  a-b-\_a-(b-c)-c].  OR-^/^  '   ^ 

2.  m  —  \n  —  (p  —  w)]. 

3.  2x-jy  +  [4*-0  +  2z)]g. 

4.  3a-{2d-[5c-(3a  +  5)]j. 

5.  a—\b  +  [c  —  (d-b)  +  (i]-2b\. 

6.  3aB-[9-(2oj  +  7)  +  3a;]. 

7.  2x-[ij-(x-21/)l 

8.  a -[2o  4- (3c -2  5)  4- a]. 

9.  (a  —  x  +  2/)  —  (b  —  a?  —  y)  +  (a  +  6  —  2  y). 

10.  3  a  -  [-  4  &  f  (4  a  -  b)  -  (2  a  -  5  tjj 

11.  4  0  -  [a  -  (2  ft  -  3  c)  +  c]  +  [a  -  (2  &  -  5  c  -  a)]. 

12.  «  +  (y- *)  -  [(3*  -2y)  +  *]+[>-<?  -  2«)]. 

13.  a-[2a+(a-2a)4  2<7,]-5a-J6a-[(a+2a)-a]5 

14.  2  a  -  (3  y  +  s)  -  {ft  -  (c  -  b)  4-  e  -  [a  -  (e  -  ft)jjf. 


15.    a  — lb -\-c 


(a  +  J)-c]  +  (2a-Hc). 


Note.  Remember  that  the  sign  —  which  is  written  in  the  last 
problem  before  the  first  term  6  under  the  vinculum  is  really  the  sign 
of  the  vinculum,  —  b  4  c  meaning  the  same  as  —  (b  4  c). 


lx  —  (x  —  5  —  *)]}< 


16.  10-ic-  j- 

17.  2x-\2x  +  (y  —  «)—  3*  +  [2oj— (y  — «— 2y)— 3«]4-4yJ. 


18.  a  -  {ft  -  [-  c  +  a  -  (a  -  J)  -  »]}  4-  [2  a  -  (J  -  a)]. 

19.  a  —  j  &  —  [a  —  (c  —  b)  +  c  —  a  - 

20.  5a-j-3a-[3a-(2a-a 

21.  20  -  a-  \7  a  -  [8  a  -  (9  a-  3  -6a)]J. 

22.  a-  j5y-[>-(3y-2«)  +  «-(a;-2y-*)]} 


(a  —  6  —  c)  —  a]  4-  ftf. 
!)-;«]  4r#}. 


56  ADDITION  AND  SUBTRACTION. 


Exercise  19. 

In  each  of  the  following  expressions  enclose  the  last  three 
terms  i*.  a  parenthesis  preceded  by  the  sign  — ,  remembering 
that  the  sign  of  each  term  enclosed  must  be  changed : 

1.  2a-b  +  3c-k  +  3e  +  5f) 

2.  x  —  a  —  y  —lb  -\-  z  -f  c  J 

3.  a  +  b-(c^±afbTl) 

4.  ax ,-f-  by  -f  cz  +  ibx  —\cy  -f-  cz\ 

5.  3a  +  2b  +  2c-[&d\-  3e  +  4/>) 

6.  x  —  y  -f  z  —  [5  xy  -\-  4  xz  4^3  ?/«.) 


Considering  all  the  factors  that  precede  x,  y,  and  z,  respectively, 
as  the  coefficients  of  these  letters,  we  may  collect  in  parentheses  the 
coefficients  of  se,  y,  and  z  in  the  following  expression  : 

ax  —  by  +  ay  —  az  —  cz  +  bx  =  ax  +  bx  +  ay  —  by  —  az  —  cz 

=  (a  +  b)x  +  (a  —  6)y  —  (a  +  c)  z. 

In  like  manner,  collect  the  coefficients  of  x,  y,  and  z  in 
the  following  expressions : 

7.  ax  -\r  by  -\-  cz  -\- bx  —  cy  -f-  az. 

8.  ax  +  2  ay  +  4  az  —  bx  -f  3  y  —  3  bz  —  2  z. 

9.  ax  —  2  by  —  5  cz  —  4  £x  +  3  cy  —  7  az. 

10.  ax  +  3  ay  +  2  by  —  bz  —  11  ex  +  2  cy  —  cz. 

11.  4  £y  —  3  ax  —  6  cz  +  2  foe  —  7  ox  —  5  cy  —  x  —  y  —  z. 

12.  6  az  —  5  by  +  3  cz  —  2  5z  —  3  ay  +  z  —  a  +  y. 

13.  z  —  6y  +  3  az  —  3  cy  -f  2  ax  —  2  HUB  —  5  &z. 

14.  x  +  ay  —  az  —  acx  -f-  &cz  —  mny  —  y  —  z. 

15.  2  ax  —  6  ay  +  46z  —  4  foe  —  2  ex  —  3cy. 

16.  ax-foc  +  2ay  +  3y  +  4az-3£z-2z. 

17.  ax  —  2  £y  -f  5  cz  —  4  foe  —  3  cy  +  az  —  2  ex  —  ay  +  4  £z. 

18.  12^x  +  12  j/y  +  4*y  -  12 5z  -  l£^cx  4-  6<jy  +  3cz. 


ADDITION  AND  SUBTRACTION.  57 

Exercise  20.  —  Eeview. 

1.  Add  4a8-  5a2-  5a*2  +  6a2*;    6  a3  +  3*8  -f  4az2  ; 
19ax2  —  llx2  —  15a2x;  10  *3  +  7  a2*  +  5  a8  — 18  ax*. 

2.  Add   3a6  +  3a  +  66  —  462;    ab  +  2a  +  45  +  9b2 ; 
7a&  —  4a  —  86  +  13a2;  6a  +  12£  —  2  a£  — 11  a2. 

Note.  Similar  compound  expressions  are  added  in  precisely  the 
same  way  as  simple  expressions,  by  finding  the  sum  of  their  coeffi- 
cients.    Thus,  3  (x  —  y)  +  5  (x  —  y)  —  2  (x  —  y)  =  6  (x  —  y). 

3.  Add4(5-*);    6(5  -«)j   3  (5  -  x) ;    -12(5 -a); 
,2(5-*);    -9(5-*).    (/(J^    ' 

4.  Add  (a  -f  5)*2  +  (b  +  c)y2  +  (a  +  c)z2-,   (b  +  c)*2  + 
(a  +  c)y2+(a  +  £)z2;  (a  +  c)x2+(a  +  &)y2-f-(&  +  c)*2. 

5.  Add    (a  +  b)x  -f-  (b  +  c)  y  -f  (c  +  a)s;   (5  +  c)z  + 
(c  +  a)x  —(a  +  b)y )  (a  +  c)  y  +  (a  +  b)  z  —  (b  +  c)x. 

6.  From  a3  —  x2  take  a3  +  2  a*  -f-  **• 

7.  From  3  a2  +  2  ax  -f-  *2  take  a2  —  a*  —  a;3. 

8.  From8*2-3a*  +  5  take  5  *2  +  2  a* -f  5. 

9.  From  a3  -f-  3  b2o  +  ab2  —  abc  take  a£2  —  abc  +  6s. 

10.  From  (a  -{-  b)  x  -\-  (a  -\-  c)  y  take  (a  —  h)x  —  (a  —  c)  y. 

11.  Simplify  7a-  J3a-  [4a-  (5a-  2a)]J.    -  S^£ 

12.  Simplify  3a-^a  +  5-[a  +  5  +  c-(a  +  5  +  c  +  cT)]5. 

Bracket  the  coefficients,  and  arrange  according  to  the 
descending  powers  of  * :  . 

13.  x2  —  ax  —  c2x2  —  bx  -f-  bx3  —  ex2  -f-  a2*8  •'"—  a8  —  ex. 

14.  2a*-3&*2-7cz3-25x  +  2c*2  +  8a*3-2e*-a*2  — foe8 

If  a  =  1,  b  =  3,  c  =  5,  and  d  =  7,  find  the  value  of : 

15.  a2  -  (£2  -  c2)  -  [J2  -  (c2  -  a2)]  +  [c2  -  (&2  -  a2)]. 

16.  a  -  25  -  \S  c  -  d—  [3  a  -  (5  5  -  e  -  8  <*)]  -  2  & j. 


58  ADDITION  AND  SUBTRACTION. 

17.  From   2  d  +  11  a  +  10  b  —  5  c'  take   2c  +  5a-3£; 

and  show  that  the  result  is  numerically  correct  when 
a  =  1,  b  =  3,  c  =  5,  d  =  7. 

18.  If  a  —  1,  b  =  —  3,   c  =  —  5,   d  =  0,  find  the  value  of 

a2  +  2b2  +  3c2  +  ±d2. 

If   a  —  3,    J  =  4,    c  =  9,    and   2s=a  +  ^  +  c,    find   the 
value  of: 

19.  s  (s  —  a)  (s  —  b)  (s  —  c). 

20.  s2  +  0  -  a)2  +  (s  -  J)2  +  (s-  c)\ 

21.  s2  -  (5  —  a)  (s  -  b)  -  (s  -  b)  (s  -  c)  -  (s  -  c)  (s  -  a). 

22.  Iix  =  a  +  2b-3c,  y  =  b  +  2c-3a,  z  =  c  +  2a  —  3b, 
show  that  x  +  y  -f-  z  =  0. 

23.  Ifx  =  a-2£  +  3c,  y  =  6-2c  +  3a,  s  =  c-2a  +  36, 

show  that  £c-f-2/  +  2  =  2a-j-2&-f-2c. 

24.  What  must  be  added  to  x2  +  5  y2  -f  3  z2  in  order  that 

the  sum  may  be  2  y2  —  22  ? 

25.  What  must  be  added  to  5  a3  —  7  a2b  +  3  ab2  in  order 

that  the  sum  may  be  a3  —  2  a26  —  2  ab2  +  #3  ? 

If  E  =  5a3  +  3a2b  -2b3,  F=3a3-  7  o?b  -  b3, 
G  =  2a2b-a3-b3,       H=  a2b -2a3 -3b3, 

find  the  simplest  expression  for : 

26.  E+\F-G-H\.  30.  E+G-(F-H). 

27.  E-\F+-G-H\.  31.  F-H-(E+G). 

28.  J£+^-G  +  J3"g.  32.  H-E-(F-G). 

29.  ^-^-(^-Jr^.  33.  F-G-(E-H). 

34.   From  2  z2  -  2  y2  -  »a  take  3  y2  +  2  x2  -  z2,  and  from 

the  remainder  take  3  z2  —  2  y2  —  x\ 
36.    Take  the  sum  of  a?c  —  2  a*  +  2  ac2  and  a8  —  ac*  —  a*c 

from  a8  —  2  a?c  +  3  ac\ 


CHAPTER  V. 
MULTIPLICATION  AND  DIVISION. 


Multiplication  of  Compound  Expressions. 

96.  Degree  of  a  Term.  A  term  that  has  one  letter  is  said 
to  be  of  the  first  degree  ;  a  term  that  is  the  product  of  two 
letters  is  said  to  be  of  the  second  degree  ;  and  so  on. 

97.  Degree  of  a  Compound  Expression.  The  degree  of  a 
compound  expression  is  the  degree  of  that  term  of  the 
expression  which  is  of  the  highest  degree. 

Thus,  a2x2  +  bx  +  c  is  of  the  fourth  degree,  since  a2x2  is  of  the 
fourth  degree. 

98.  When  all  the  terms  of  a  compound  expression  are 
of  the  same  degree,  the  expression  is  said  to  be  homogeneous. 

Thus,  xs  +  3  x2y  +  3  xy2  +  y*  is  a  homogeneous  expression,  every 
term  being  of  the  third  degree. 

99.  Dominant  Letter.  If  there  is  one  letter  in  an  expres- 
sion of  more  importance  than  the  rest,  it  is  called  the 
dominant  letter ;  and  the  degree  of  the  expression  is  called 
by  the  degree  of  the  dominant  letter. 

Thus,  a?x2  +  bx  +  c  is  of  the  second  degree  in  x. 

100.  Arrangement  of  a  Compound  Expression.  A  compound 
expression  is  said  to  be  arranged  according  to  the  powers 
of  some  letter  when  the  exponents  of  that  letter  descend  or 
ascend,  from  left  to  right,  in  the  order  of  magnitude. 

Thus,  3  ax3  —  4  bx2  —  6  ax  +  8  b  is  arranged  according  to  the  de- 
scending powers  of  x ;  and  86  —  6  ax  —  4  bx2  +  3  ax3  is  arranged 
according  to  the  ascending  powers  of  x. 


60  MULTIPLICATION  AND  DIVISION. 


Multiplication  of  Polynomials  by  Monomials. 

We  have  a  (b  +  c)  =  ab  +  ac ;  (§  41) 

and  a(b  —  c)  =  ab  —  ac.     Hence, 

101.   To  Multiply  a  Polynomial  by  a  Monomial, 

Multiply  each  term  of  the  polynomial  by  the  monomial, 
and  connect  the  partial  products  with  their  proper  signs. 

Find  the  product  of  ab  +  ac  —  be  and  abc. 

ab  -\-  ac  —  be 
abc 


a2b2c  +  a2bc2  -  ab2c* 


Note.     We  multiply  ab,  the  first  term  of  the  multiplicand,  by  abc, 
and  work  to  the  right. 

Exercise  21. 

i 

Multiply : 

1.  5a  +  3b  by  2a2.  7.  a2  +  b2  -  c2  by  a3bc2. 

2.  ab  -  be  by  5  a3bc.  8.  5  a2  -  3  b2  +  2  c2  by  Aabsc2. 

3.  aft  —  ac  —  be  by  a&c.  9.  abc  —  3  a36c2  by  —  2  ab2c. 

4.  6  a5&  -  7  a262c  by  a262c.  10.  xyz2  +  ScV*  by  -  *V- 

5.  cc  —  ?/  —  z  by  —  3xBy7z9.  11.  3x  -  2  ?/ —  4  by  5a2. 

6.  x2  +  2if-zby  -3x2.  12.  3x2  -  4  y2  +  5s2  by  2arfy. 

13.  a3x  —  5  a2#2  -f  #£3  +  2  a4  by  ax2y. 

14.  -  9  a5  +  3  a362  -  4  a2bs  -  b5  by  -  3  a&4. 

15.  3z3-2z2?/-7£c?/2  +  ?/3by  -5x2y. 

16.  -4V  +  5x2//  +  8a;3  by  -3  a2?/. 

17.  -  3  .+  2  ab  +  a2^2  by  -  a4. 

18.  -  *  -  2  a*2  +  5  afyt;2  by  -  3  x*yz. 


MULTIPLICATION  AND  DIVISION.  61 

Multiplication  of  Polynomials  by  Polynomials. 

If  we  have  m  +  n  -f  p  to  be  multiplied  by  a  -f-  J  4-  c,  we 
may  substitute  Jf  for  the  multiplicand  m  +  K+f     Then 

(a  +  b  +  c)  Jf  =  aM  +  bM+  cM.  (§  41) 

If  now  we  substitute  for  M  its  value  m  -±-n+p,we  have 

aM  +  &Jf-f  cilf 

=  a  (m  -f-  w  +p)  +  b(m  +  n  -\-p)  +  c(m  +  n  +i?) 
=  awi  -f-  aw  +  ap  +  #w  +  #ra  +  bp  -f-  cm  -f-  en  +  cjp. 

102.  To  Find  the  Product  of  Two  Polynomials,  therefore, 
Multiply  every  term  of  the  multiplicand  by  each  term  of 
the  multiplier,  and  add  the  partial  products.  « 

In  multiplying  polynomials,  it  is  a  convenient  arrange- 
ment to  write  the  multiplier  under  the  multiplicand,  and 
place  like  terms  of  the  partial  products  in  columns. 

1.   Multiply  5a-  6b  by  3a-  4& 
5a  —    6b 
Sa  -    4fr 
15  a2  -  18  ab 

-20ab  +  2±b2 
15  a2  -  38  ab  +  24  b* 

We  multiply  5  a,  the  first  term  of  the  multiplicand,  by  3  a,  the  first 
term  of  the  multiplier,  and  obtain  15a2;  then  we  multiply  —  66,  the 
second  term  of  the  multiplicand,  by  3  a,  the  first  term  of  the  multi- 
plier, and  obtain  —  18  ab.  The  first  line  of  partial  products  is  15  a2 
—  18  ab.  In  multiplying  by  —  4  6,  we  obtain  for  a  second  line  of 
partial  products  —  20  ab  +  24  62,  and  this  is  put  one  place  to  the  right, 
so  that  the  like  terms  —  18  ab  and  —  20  ab  may  stand  in  the  same 
column.  We  then  add  the  coefficients  of  the  like  terms,  and  obtain 
the  complete  product  in  its  simplest  form. 


62  MULTIPLICATION  AND  DIVISION. 

2.   Multiply  4  a;  +  3  +  5  a;2  -  6  a;3  by  4  -  6x2  -  5x. 

Arrange  both  multiplicand  and  multiplier  according  to 
the  ascending  powers  of  x. 

3+    4a;  +    5x2-    6a;8 
4-    5x-    6x2 


12  +  16a;  +  20a;2  -24a;8 

-  15a;  -  20a;2  -  25a;8  -f  30a;4 

-  18  a;2  -  24  a;8  -  30  a;4  +  36  a;* 

12+      x  -18a;2  -73a;3  +  36a^ 


3.   Multiply  1  +  2a;  +  a;4  -  3x2  by  x3  -  2  -  2x. 

Arrange  according  to  the  descending  powers  of  x. 

x*-3x2  +  2x  +1 
x8-2x  -2 


-2x5  +  6a;8 -4a;2 -2a; 

-2a;4  +6a;2-4a;-2 

-5a*  +7a;3  +  2a;2-6a;-2 


4.   Multiply  a2  +  b2  +  c2  —  ab  —  be  —  ac  by  a  -+-  b  -f-  c. 

Arrange  according  to  the  descending  powers  of  a. 

a2  —  ab  —  ac  -f-    62  —       5c  +    c2 
a  +     #  +     £ 


a8  —  a2£  —  a2c  -f-  a&2  —     afo  +  ac2 

+  a2b  -ab2-     abc  +b8-b2c  +  bc2 

+  a2c —     abc  —  ac2         +  b2c  —  be2  +  cz 

'a1  -3  abc  +b*  +? 

Note.  The  student  should  observe  that,  with  a  view  to  bringing 
like  terms  of  the  partial  products  in  columns,  the  terms  of  the  multi- 
plicand and  multiplier  are  arranged  in  the  same  order. 


MULTIPLICATION  AND  DIVISION.  63 

Exercise  22. 

Multiply : 

1.  x  +  10  by  x  +  6.  12.  2* -3  by  a; +  3. 

2.  x  —  2  by  x  -  3.  13.  a;  -  7  by  2  a;  -  1. 

3.  se  —  3  by  x  +  5.  14.  m  —  n  by  2  m  +  1. 

4.  a;  +  3  by  a;  —  3.  15.  m  —  a  by  m  +  a. 

5.  a;  -  11  by  a;  -  1.  16.  3a;  +  7  by  2a;  -  3. 

6.  -a; +  2  by  -a;- 3.         17.  5x  -  2y  by  5x  +  2y. 

7.  —x-  2  by  x  —  2.  18.  3a;  -  4y  by  2a;  +  3y. 

8.  —  a;  +  4  by  x  —  4.  19.  a;2  +  y2  by  a;3  —  y3. 

9.  -  a;  +  7  by  a;  +  7.  20.  2  x2  +  3  y2  by  x2  +  y2. 
10.  a;  —  7  by  a;  +  7.  21.  x-\-y-\-zbyx  —  y-\-z. 
LI.  a;-3by2a;  +  3.  22.  a;  +  2y-s  by  a;  —  y  +  2s, 

23.  x2  —  xy  +  y2  by  a;2  +  a;y  +  y2. 

24.  m2  —  mn  +  w2  by  m  +  w. 

25.  m2  +  mn  +  n2  by  m  —  n. 

26.  a2  -  3  ab  +  b2  by  a2  -  3  a£  -  b2. 

27.  a2 -7a +  2  by  a2  -  2  a  +  3. 

28.  2x2-3xy-{-4,y2bj  3x2  +  4:xy-5y2. 

29.  a;2  +  xy  +  y2  by  x2  —  xz  —  z2. 

30.  a;2  4-  y2  +  22  —  a;y  —  xz  —  yz  by  a;  +  y  +  *c 

31.  ±a2-10ab  +  25a2b2by  5b  +  2a. 

32.  a;2  +  4  y  by  y2  +  4  x. 

33.  a;2  +  2a;y  +  8  by  ?/2  +  2a;y-8. 

34.  a2  +  J2  +  l-a5-a-&bya  +  l+S. 

35.  3a;2-2y2  +  5s2by  8a;2  +  2i/2-3«2. 

36.  a2  +  b2  -{-  c2  —  ab  —  ae  —  bo  by  a  -{-b  +  c. 

37.  x2  —  xy  +  yt  +  x  +  y  +  lbyx  +  y  —  l. 


64  MULTIPLICATION  AND  DIVISION. 

38.  5x  +  4:X2  +  x8-2±byx2  +  ll-4:X. 

39.  x8  +  llx-4;x2-24:by  x2  +  5  +  4x. 

40.  x4  +  x2-4:x-ll+2xsby  x2-2x-\-3. 

41.  -5x4~x2-x-\-xs-\-13x3by  x2-2-2x. 

42.  3x  +  x8-2x2-±by  2x  +  4x3  +  3x2 +  1. 

43.  5a44-2a2^24-^8-3a3^  by  5a8b  -  2  ab3  +  3  a2b2  +  b*. 

44.  4a7y-32a?/4-8ay  +  16ay  bjay  +  4ay  +  4«y. 

45.  6a5b  +  3  a2b4  -  2  ab5  +  61  by  4  a4  -  2  a&8  -  3  J4. 

46.  ic2  +  2/2  +  2  ay  -  2  a  -  2  ?/  -  1  by  x  +  y  —  1. 

47.  am  +  2  a"1'1  —  3  am~2  -  1  by  a  +  1. 

48.  an  -  4  a""1  +  5  an~2  +  an~3  by  a  -  1. 

49.  a4n+1  —  4  a3n  +  2  a2*"1  —  an~2  by  2  a3  -  a2  +  a. 

50.  aJ*  —  yn+1  by  xn  -f  yn+1. 

51.  a2"  +  2  ay1  +  y2"  by  x2n  —  2  xny*  +  y2n. 

52.  an  -  an+1  -  an+2  +  1  by  an  +  1. 

53.  a8n  —  a2n  +  an  —  l  by  an  —  1. 

54.  an  —  a""1  +  an~2  +  an~3  by  a2  -  1. 

Exercise  23.  —  Review. 
Simplify : 

1.  (x2  -x  -19)  (x2  +  2x-3). 

2.  (l  +  2a-f  a2)(l-zs4-2z2-3a;). 

3.  (2z24-2-f  3x)(2a-3a24-2arJ). 

4.  (3a2  +  5-4ic)(8  +  6a;2-7ic). 

5.  (x8  4-  #  —  y)  —  (ic2  —  y2  4-  ay)  (a  —  1). 

6.  (3  4-  7X2  -  5x)  (8 x2-  6x  -10sc3-f  4). 

7.  (b8  +  6  ab2  -  4  a2£)  (2  a2£  -  ab2  -  8  a8). 

8.  (x2  +  ax-b)-(x2  +  5x-±)(x-3). 

9.  (a^  —  wa;2  +  nx  +  r)  (x2  -\-  ex  +  d). 
10.  Jx2-(a4-5)a;4-^Ka;-c)- 


MULTIPLICATION  AND  DIVISION. 


11.  (xs  +  x*y  +  xy2  +  y*)(x-y). 

12.  (Ax2  +  9y2-  6x7/)  (4z2  +  §tf  +  §xy). 

13.  (z4-3a;2  +  5)(a2  +  4)(a;-2).  ? 

14.  (z4  -  *%»  +  y*)  (x*  +  x2f  +  y4). 

15.  (2a5-3z3  +  4z2-5)(a;2-8)(a;-3).     ' 

16.  (a2m  —  amym  +  y2™)  (am  +  ym)(a  +  y)> 

17.  (aSx  -  a2x  +  ax  -  1)  (ax  +  1)  (ax  -  1). 

18.  (a2  +  b2  +  c?-h2ab-ac-bc)(a  +  b  +  c). 

19.  (a-2b)(b-2a)-(a-3b)(±b-a)  +  2ab.    — 

20.  (a  +  &  +  c)  (a  +  b  -  c)  -  (2  ab  -  c2). 

21.  (m  +  ri)m  —  [(w  —  ri)2  —  n(n  —  m)~].     — 

22.  [ac-(a-ft)(ft4-c)]-ft[ft-(a-c)]. 

23.  (a  -  1)  (a*  -  2)  -  3x(x  +  3)  +  2 [(a;  +  2)  (i  +  1)-  3]. 

24.  4  (a  -  3  5)  (a  +  3  J)  -  2  (a  -  6  ft)2  -  2  (a2  +  6  ft2). 

25.  (x-\-y  +  z)2—x(y  +  z  —  x)  —  y(x+z  —  y)—z(x-}-y--z). 

26.  5  j  (a  —  J)  a;  —  c?/  \  —  2  \  a  (x  —  y)  —  bx  \  —  (3  ax  —  5  cy). 

27.  (a  -  b)  x  -  (ft  -  c)  a  -  j  (ft  -  x)  (ft  -  c)  -  (ft  -  c)  (ft  +  c)  \. 

28.  (a  +  b)  (b  +  c)  —  (c  +  d)  (a  +  d)  -  (a  +  c)  (ft  —  d). 

29.  a2  (b  -  c)  -  b2  (a-  c)  +  <?  (a-b)-(a-  b)(a  -c)(b-  c). 

30.  (2  a  -  b)2  +  2  b  (a  +  b)  -  3  a2  -  (a  -  J)2  +  (a  +  ft)  (a  -  ft). 
If  a  =  0,  J  =  1,  c  =  —  1,  find  the  value  of: 

31.  (b-c)2+(c-a)2  +  (a-b)\       ' 

32.  (b  +  cY-(o-a)2-(a-b). 

33.  (a-b)(a-c)  +  c(3a-b-c)  +  2ac-(a-c)  +  2b. 

34.  (ft  -  c)8+  (a  -  ft)3  +  (c  -  a)*-  3  (ft  -  c)  (a  -  ft)  (c  -  a). 

35.  If  a  =  12,  and  ft  =  5,  find  the  values  of 
V^~2+  Vft*:  and  Va2  +  ft2. 


66  MULTIPLICATION  AND  DIVISION. 

Division  of  Compound  Expressions. 
Division  of  a  Polynomial  by  a  Monomial. 
By  the  distributive  law,  (§  43), 

9a4b2x-  12  a3bx2-3a2x      9  a4b2x      12  asbx2      3a2x 


1. 


3  a2x  3  a2x  3  a2x         3  a2# 

=  3  a2b2  -±abx-  1. 


2*2*-1         ""2  a;2-1       2^-1_^  ^*     ' 

Note.  Here  we  have  4  n  + 1  —  (2n  —  l)  =  4n+l—  2n+l  =  2n  +  2, 
and  3n—  (2n  —  l)  =  3n  —  2n  +  l  =  n  +  l,  as  indices  of  x  in  the  first 
and  last  terms  of  the  quotient  respectively. 

103.   To  Divide  a  Polynomial  by  a  Monomial,  therefore, 
Divide  each  term  of  the  dividend  by  the  divisor,  and  con- 
nect the  partial  quotients  with  their  proper  signs. 

Exercise  24. 
Divide : 

1.  x2  +  2xyhyx.  3.    4  xG  -  8  x3  by  2  xs. 

2.  a2-2abbja.  4.    -  6x3  -  2x  by  -  2x. 

5.  -  8  a5  -  16  a10  by  -  8  a3. 

6.  27  a3  -  36  a2  by  9  a2. 

7.  -  30  a7  +  20  a9  by  -  10  a6. 

8.  -12:ry-4xy  by  -4<bY- 
9.-3  arV  -  6  #V  by  -  3  af«*. 

10.  3  aW  --  9  aW  by  3  aW. 

11.  x2  —  xy  —  xz  by  —  x. 

12.  3a8-6a26-9a52by  -3  a. 

13.  #5?/2  —  ay  —  x2y2  by  a2?/2. 

14.  a%2c  -  a2bsc  -  a2bc*  by  abc. 

15.  8a8--4a2£-6a&2by  -2a.- 

16.  5  mzn  —  10  m2n2  —  15  wwi8  by  5  m». 


MULTIPLICATION  AND  DIVISION.  67 

Division  of  One  Polynomial  by  Another. 
If  the  divisor  (one  factor)       =  a  -f-  b  -f-  c, 

and  the  quotient  (other  factor)  =  n-\-p  +  g, 

(  +  an  +  bn  +  en 
then  the  dividend  (product)       —  <  +  ap  -\- bp -\-  cp 

[  +  aq  -f  hq  +  c^. 

The  first  term  of  the  dividend  is  an ;  that  is,  the  product 
of  a,  the  first  term  of  the  divisor,  by  n,  the  first  term  of  the 
quotient.  The  first  term  n  of  the  quotient  is  therefore 
found  by  dividing  an,  the  first  term  of  the  dividend,  by  a, 
the  first  term  of  the  divisor. 

If  the  partial  product  formed  by  multiplying  the  entire 
divisor  by  n  is  subtracted  from  the  dividend,  the  first  term 
of  the  remainder  ap  is  the  product  of  a,  the  first  term  of  the 
divisor,  by  p,  the  second  term  of  the  quotient ;  that  is,  the 
second  term  of  the  quotient  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  first  term  of  the  divisor. 
In  like  manner,  the  third  term  of  the  quotient  is  obtained 
by  dividing  the  first  term  of  the  new  remainder  by  the 
first  term  of  the  divisor ;  and  so  on.     Therefore, 

104.   We  have  the  following  rule  for  division  : 

Arrange  both  the  dividend  and  divisor  in  ascending  or 
descending  powers  of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor. 

Write  the  result  as  the  first  term  of  the  quotient. 

Multiply  all  the  terms  of  the  divisor  by  the  first  term  of 
the  quotient. 

Subtract  the  product  from  the  dividend. 

If  there  is  a  remainder,  consider  it  as  a  new  dividend  and 
proceed  as  before. 


68 


MULTIPLICATION  AND  DIVISION. 


105.  It  is  of  fundamental  importance  to  arrange  the  divi- 
dend and  divisor  in  the  same  order  with  respect  to  a  com- 
mon letter,  and  to  keep  this  order  throughout  the  operation. 

The  beginner  should  study  carefully  the  processes  in  the 
following  examples : 

1.   Divide  x2  +  18  x  +  77  by  x  +  7. 


x2  +  18  x  +  77 
a?+    7x 


x  +  1 


x  + 11 


11  a +  77 
11* +  77 


Note.     The  student  will  notice  that  by  this  process  we  have  in 

effect  separated  the  dividend  into  two  parts,  x2  +  7x  and  11  x  +  77, 

and  divided  each  part  by  x  +  7,  and  that  the  complete  quotient  is  the 

sum  of  the  partial  quotients  x  and  11.     Thus, 

x2  +  18  x  +  77  =  a*  +  7x  +  Its  +  77  =  (x2  +  7x)  +  (llx  +  77); 

x2  +  18x  +  77     x2  +  7xJllx  +  77         ,  in 
-I rir~  =  x  +  11. 


x  +  7 


x  +  7 


x  +  7 


2.   Divide  a2  -  2  ab +  b2  by  a- b. 


a2-2ab  +  b2 

a  —  b 

a2  —     ab 

a  —  b 

-  ab  +  b2 

-  ab  +  b2 

3.    Divide  4  aAx2  -  4  a2xA  +  x6  -  a«  by  x2  -  a\ 

Arrange  according  to  descending  powers  of  x. 

xe  -  4  a2x*  +  4  aV  - 


<c4  -  3  a  V  +  a4 


-3aV  +  4aV 
-3«¥  +  3a¥ 


a¥ 


MULTIPLICATION  AND  DIVISION. 


69 


4,   Divide  22  aV  4  15  b*  4  3  a4  -  10  a3J 

Arrange  according  to  descending  powers 
3  a4  -  10  tf*  +  22  a2b2  -  22  aft3  4  15  b4 

-22ab8 

'  a2  +  3b2-2ak 
of  a. 
a2-2ah  +  3b'k 

3  a4-    6a3&4    9^2 

3a2-±ab  +  hb'k 

-  4»%  +  13  a262  -  22  a&3  4  15  ft4 

-  4a8H    8a262-12a&8 

5a262-10a&3  +  15&4 
5  a262  -  10  «&3  +  15  b* 

5.   Divide  5x8  -  x +  1  -  Sx4  hj  1  +  3x2  -  2x. 
Arrange  according  to  ascending  powers  of  x. 


1-     x  +  5x3-3x4 
l-2x-\-3x2 


1  -  2  x  +  3  a2 


14-    ic  —     sca 


z-3a2  +  5cc3 

-3z4 

»  -  2  z2  4  3  xs 

-     «2  +  2a;8-3x4 

-     z2  +  2tf8-3a;4 

6.   Divide  x3  -{-  y3  +  z3  —  3  xyz  hy  x  +  y  +  z. 
Arrange  according  to  descending  powers  of  x. 


xz  —  3  xyz  4  y8  +  z3 
x3  -f-  x2y  4-  #2^ 


a;  +  y  4  « 


—  x"y  —  ar« 

q;2y  —  xy2  —    xyz 


1  —  xy  —  xz  +  y2  —  yz  +  z* 
3xyz  +  y3  +  z3 


—  x'2z  4     xy2 

—  x2z 

-2xyz  +  f  +  z8 
—     xyz  —  xz2 

xy2 
xy2 

—     xyz  4  xz2  +  y8  +  z8 
+  f  +  f* 

—  #?/£  4  xz2  —  y2^  4  z3 

—  xyz             —  y2z  —  yz2 

xz2  4  y^2  4  £8 
«*2  4  ys2  4  z* 


70  MULTIPLICATION  AND  DIVISION. 

7.   Divide  4  ax+1  -  30  a*  -h  19  a*'1  -f  5  a*'2  +  9  ax"4 

by  a*"3  —  7  ax~4  +  2  ax~5  —  3  a*-6. 


4ax+1-30aa;4-19aiE-1+  5ax-2+9ax-4 
4ax+1-28ax-f  8a*-1-12ax-2 


aa>~s~7ax-4-\-2ax-'i-3ax-^ 


4a4-2a3-3a2 


—  2ax+llax-1+17ax-2H-9ax-4 

-  2ax+Uax-1-  4ax~2+6ax-3 

-  3ax-14-21ax-2-6ax"-3+9ax-4 

-  3^x-1+21ax-2-6ax-8+9^x-4 

Note.  We  find  the  index  of  a  in  the  first  term  of  the  quotient  by 
subtracting  the  index  of  a  in  the  first  term  of  the  divisor  from  the 
index  of  a  in  the  first  term  of  the  dividend.  Now,  (x  +  1)  —  (x  —  3) 
=  x  +  1  —  £  +  3  =  4.  Hence,  4  is  the  index  of  a  in  the  first  term  of 
the  quotient.     In  the  same  way  the  other  indices  are  found. 

Exercise  25. 
Divide : 

1.  a2  +  la  +  12  by  a  +  4.  6.  4cc2  +  12  x  +  9  by  2  x  +  3. 

2.  a2  -  5  a  +  6  by  a  -  3.  7.  6  a2  -  11  a;  +  4  by  3  a  -  4. 

3.  a2  4- 2  a 7/ +  ?/2  by  aj  +  y.  8.  8  z2-10  aa-3  a2  by  4  cc+a, 

4.  x2  —  2  xy  -\-  y2  by  x  —  y.  9.  3  a2  —  4  a  —  4  by  2  —  a. 
B.  x2  —  y2  by  x  —  y.  10.  a3  —  8  a  —  3  by  3  —  a. 

11.  a4  +  11  a2  -  12  a  -  5  a3  +  6  by  3  +  a2  -  3  a. 

12.  y4-  9f  +  y*  -lGy  -4  by  ^  +  4  +  43/. 

13.  36  +  m4  -  13  m2  by  6  -f  m2  +  5  m. 

14.  l-s-3s2-s6byl+2s  +  s2. 

15.  £6  -  2  58  -f- 1  by  £2  -  2  J  +  1. 

16.  ar*  +  2a-Y  +  9y*by  x2  -  2 ary  +  3 ^. 

17.  a5  +  £5  by  a4  -  a*b  +  «2&2  -  abs  +  **. 

18.  1  +  5ar*  -  6z4  by  1-  x  +  3a* 


MULTIPLICATION  AND  DIVISION.  71 

19.  8zY  +  92/4-|-16x4by  4x2  +  3y2  -  4=xy. 

20.  xz  +  tf +  z3  +  3x2y +  3xy2  by  x  +  y  +  z.     ' 

21.  as  +  bs  +  c*-3abcby  a  +  b  +  c. 

22.  a3  +  8y8  +  «3-6xys  by  x2  +  ±tf+ z2 -xz  —  2xy-2yz. 

23.  2a2  —  3y2  +  a;?/  —  jb*  —  4=yz  —  z2  by  2x +  3y  +  z. 

24.  x2  —  y2  — 2 yz  —  z2  by  x-\-y-\-  z. 

25.  a;4  +  jc2^2  +  y4  by  a;2  +  xy  +  2Z2. 

26.  a:4-9a:2  +  12a;-4by  x2  +  3x-2. 

27.  ^-22/4-6^  +  42/2  +  13y  +  6by  2/3  +  3^  +  3y  +  l. 

28.  2/4-5?/2«2  +  4^by  y2-yz-2z2. 

29.  a:2-4?/2-9s2  +  12yzby  a;  +  2?/--3z. 

30.  a:5 -41a; -120  by  x2  +  ±x  f  5. 

31.  cc4-3-f  5z  +  z3-4z2by  3-2a:-a;2. 

32.  6-2z4-f  10a3-lla:2  +  a;by  4Z-3-2X2. 

33.  l-6x5  +  5x6by  l-2x-^x2. 

34.  a:4 +  81  + 9a;2  by  3a; -a2 -9. 

35.  xz  —  y3  by  x2  +  xy  +  2Z2- 

36.  a;6  +  /  by  x2  +  y2. 

37.  a;4  +  a2x2  +  a4  by  x2  -  ax  +  a2. 

38.  a2  -  2  b2  +  a&  -  3  c2  +  1  be  +  2ac  by  3  c  +  a  -  &. 

39.  ab  +  2a2  -  3 b2  -  46c  -  ac  -  c2  by  c  +  2a  +  3b. 

40.  15  a4  +  10a3x  +  ±a2x2  +  6axs  -  3x4by  3a2  +  2ax- x2 

41.  as-8b3-l-6abby  a-2b-l. 

42.  0^  —  3  a;2nyn  +  3  xny2n  —  y8"  by  xn  —  yn. 

43.  am+nbn  —  4  am+n-1&2n  —  27  am+n-2b3n  +  42  a»»+»-3&*» 

by  am  +  3  a"1"1^  —  6  a7"-2^". 


72  MULTIPLICATION  AND  DIVISION. 

106.  Integral  expressions  may  have  fractional  coefficients, 
since  an  algebraic  expression  is  integral  if  it  has  no  letter 
in  the  denominator.  The  processes  with  fractional  coeffi- 
cients are  precisely  the  same  as  with  integral  coefficients, 
as  will  be  seen  by  the  following  examples  worked  out : 

1.  Add  £a2  -  £  ab  +  \  b2  and  |  a2  +  §  ab  -  §#. 

%a2-%ab  +  \b2 
f  a2  +  f  «&  -  f  62 
%a2  +  \ab-\b2 

2.  From  \a2-\ab  +  \b2  take  \a2-\ab  +  %b2. 

\a2-\ab  +  |P 
ja'-frq6  +  g&2 
£a2  +  £«6-TV2 

3.  Multiply  \a2-\ab  +  J&2  by  \a-%h 

\a2-%ab  +    \b2 

\a  -%b 

\az-\a2b  +    \ab2 

-ja2b+    ^ab2-^ 
±as-$a2b  +  $§ab2-%b3 

4.  Divide  f  b*  +  ^-bd2  -  $$%b2d  -  ^d*  by  }ft  -  §d. 

3*  ~$d 


g&3-    tfPd 


tV-$bd+\ffl 


ftPd  +  *+beP-ih(P 
A  62dT  4-      fa*2 


MULTIPLICATION  AND  DIVISION.  73 

Exercise  26. 

1 .  Add  |  a2b  +  J  W  +  A  and  -  ^  a2&  -}^-J. 

2.  From  f  x2  +  3  ax  —  f  a2  take  2  a2  —  §  ax  —  £  a2. 

3.  From  \y  —  %a  —  \x  +  \b  take  J y  +  *£ a  —  § x. 

4.  Multiply  Jc2-ic-^by  J^-Jc  +  J. 

6.   Multiply  \x-\x2  +  \x8by  ^x  +  ^tf  +  ^x*. 

6.  Multiply  0.5  m4  -  0.4  m8rc  +  1.2  m2w2  +  0.8  mnz  -  1.4  rc4 

by  0.4  m2  -  0.6  rarc  -  0.8  n2. 

7.  Divide  -ft  a4  -  {  a8&  +  Jf  a2£2  +  J  a&8  by  f  a  +  £&. 

8.  Divide-^6  +  ^2-|J^8  +  |^4by  -  je?  +  2d 

Exercise  27.  —  Review. 

1.  Find  the  value  of  x8  +  y8  +  *8  —  3  ai/2,  if  a;  =  1,  y  =  2, 

and  2  =  —  3. 

2.  Find  the  value  of   V2  6c  —  a,  and  of  V2  6c  —  a,  if 

6  =  8,  c  =  9,  and  a  =  23. 

3.  Add  a2b  -  ab2  +  £8  and  a8  -  £  a26  +  ab2  -  f  &8. 

4.  Multiply  a2m  —  a"1^  +  b2m  by  am  +  bm. 

5.  Multiply  4  a2m+4  +  6  a"1*8  +  9  a2  by  2  am+4  -  3  a9. 

6.  Divide  a8  +  8  y8  -  125  zs  +  30xyzby  x +  2y  -  5z. 

7 .  Simplify  (aj  -  a)2  -  (a*  -  b)2  -  (a  -  b)  (a  +  b  -  3  x). 

8.  Find  the  coefficient  of  x  in  the  expression 

x  + a- 2\2  a -b(c-  a;)]. 

9.  Multiply  4  a^+2"-1  -  7  iC2m-3»+2  +  5  x2**8™-2  by  5  a?2-"1-9". 

10.  Divide  c2^2*  -  c4^4-2*  -  c4^8-*  by  c^d2-2*. 

11.  Divide  raV  —  m1+x2/1+w  +  m5-ay*-*  by  ms-xyn^\ 


74  MULTIPLICATION  AND  DIVISION. 

12.  Divide  al+3v  -  a?  -f  a2+5»  by  a2~x+K 

13.  Divide  x  —  a;1*-*""4  -f  a;2m  by  a?2-™-2". 

14.  Divide  ^  —  %f-*>n  +  y4p+1  by  y^-**1. 

15.  Divide  2  a3n  -  6  a;2^"  +  6xny2n  —  2ySnbyxn-yn. 

16.  Divide  x8  -  2  asc2  +  a2ic  -  a&a;  -  62a;  +  a2b  +  a&2 

by  x2  —  ax  -\-bx  —  ab. 

17.  Divide  z4wi  +  x2"1  +  lby  x2™  -  xm +  1. 

18.  Divide  3  ccm+7  —  4  aTO+6  —  12  xm+5  —  9  xm+4 

by  xm+*  —  3  #m+3. 

19-   Divide  6  sc4"*5  -  13  xSm+s  +  13  a2OT+5  -  13  xm+s  -  5  x6 
by  2  sc2™  -  3  xm  -  1. 

20.  Divide  12  a5"-3  -  a4""2  -  20  a3""1  -f  19  a2n  -  10  an+1 

by  4  a2"  -  3  an+1  +  2  a2. 

21.  Arrange   according   to   descending   powers   of   x   the 

following  expression,  and  enclose  the  coefficient  of 
each   power   in   a  parenthesis   with  a  minus   sign 
before  each  parenthesis  except  the  first : 
xs  —  2  bx  —  a2x*  —  ax  —  ax2  —  ex  —  a2xz  —  bcx. 

22.  Divide  1.2  a*x  -  5.494  asx2  +  4.8  a2xs  +  0.9  ax4  -  Xs 

hy0.6ax-2x2. 

23.  Multiply  \a?-\ab±\b2\)y  %a  +  \b. 

24.  Multiply  %a2  +  ab  +  f&2by  %a-\b. 

25.  Divide  *  a8  +  T^  a&2  +  TV  &8  by  £  a  +  £  b. 

26.  Subtract  J  a2  +  £  a?y  +  £  y2  from  £  «2  -  £  xy  +\  y2. 

27.  Subtract  o^  -\-\xy  —  \y2  from  2 a-2  —  \ xy  +  y*. 

28.  If  a  =  8,  5  =  6,  c  =  —  4,  find  the  value  of 

Va2  +  2  6c  +  V62  +  ac  -f  Vc2  +  ok 


CHAPTER  VL 
SPECIAL  RULES. 


Multiplication. 

107.  Square  of  the  Sum  of  Two  Numbers. 

(a  +  by  =  (a  +  b)(a+.  b) 

=  a(a  +  b)  +  b(a  +  b) 

=  a2  +  ab  +  ab  +  b2 

=  a2  +  2ab  +  b2.        Hence, 

Rule  1.     The  square  of  the  sum  of  two  numbers  is  the 
sum  of  their  squares  plus  twice  their  product. 

108.  Square  of  the  Difference  of  Two  Numbers. 

(a  -  b)2  =(a-b){a-b) 

=  a  (a  —  b)  —  b  (a  —  b) 

=  a2  —  ab  —  ab  -\- b2 

—  a2  —  2ab  -f-  b2.         Hence, 

Rule  2.     The  square  of  the  difference  of  two  numbers  is 
the  sum  of  their  squares  minus  twice  their  product. 

109.  Product  of  the  Sum  and  Difference  of  Two  Numbers. 

(a  +  b)  (a  -  b)  =  a  (a  -  b)  +  b  (a  -  b) 
—  a2  —  ab  -\-  ab  —  b2 
=  a2-  b2.  Hence, 

Rule  3.     The  product  of  the  sum  and  difference  of  two 
numbers  is  the  difference  of  their  squares. 


76  SPECIAL  RULES   OF  MULTIPLICATION. 

110.  The  following  rule  for  raising  a  monomial  to  any- 
required  power  will  be  useful  in  solving  examples  in  multi- 
plication : 

Raise  the  numerical  coefficient  to  the  required  power, 
and  multiply  the  exponent  of  each  letter  by  the  exponent  of 
the  required  power. 

Thus  the  square  of  7  a266  is  49  a4612. 

Exercise  28. 
Write  the  product  of  : 

1.  (x  +  y)\  7.  (x  +  y)(x-  y). 

2.  (x-a)\  8.  (4*  -3)  (4z  +  3). 

3.  (x  +  2b)2.  9.  (3a2  +  4tb2)  (3a2-±b*). 

4.  (Sx  -2c)2.  10.  (Sa-c)(Sa-c). 

5.  (4y-5)2.  11.  (x  +  lb^ix  +  lb2). 

6.  (3  a2 +  4  a2)2.  12.  (ax  +  2by)(ax-2by). 

111.  If  we  are  required  to  multiply  a  +  b+cby  a  +  b  —  c> 
we  may  abridge  the  ordinary  process  as  follows : 

(a  +  b  +  c)  (a  +  b  -  c)  =  \(a  +  b)  +  c\\(a  +  b)  -  c\ 
By  Kule  3,  =  (a  +  b)2  -  c2 

By  Rule  1,  =  a2  +  2ab  +  b2  -  c2. 

If  we  are  required  to  multiply  a  +  b  —  chj  a  —  b  +  c,  we 
may  put  the  expressions  in  the  following  forms,  and  per- 
form the  operation : 

(a  +  b  -  c)  (a  -  b  +  c)  =  \a  +  (b  -  c) \\a  -  (b  -  c)\ 
By  Rule  3,  =  a2  -  (b  -  c)2 

By  Rule  2,  =  a2  -  (b2  -2bc  +  c*) 

By  §  40,  =  a2  -  b2  -f  2bc  -  c*. 


SPECIAL  RULES   OF  MULTIPLICATION.  77 

Exercise  29. 
Find  the  product  of : 

1.  x  +  y  4-  z  and  x  —  y  —  z, 

2.  x  —  y  4-  z  and  x  —  y  —  z. 

3.  owe  +  by  -f  1  and  ax  +  by  —  1. 

4.  1  +  a;  —  y  and  1  —  a;  -f  y. 

5.  a  +  2b-  3c  and  a-  2b  +  3c. 

6.  a2  -  ab  +  62  and  a2  +  ab  +  62. 

7.  ra2  4-  mw  4-  rc2  and  m2  —  m»  4-  n2. 

8.  2  +  cc  +  x2  and  2  -  x  -  x2. 

9.  a2  +  a  4- 1  and  a2  —  a  +  1. 

10.  3#  +  2y-2  and  3#-2y  +  s. 

11.  1  +  #  4-  y  and  1  4-  x  —  y. 

12.  a2-2acc4-4a;2anda2  +  2aa;4-4aj2. 

13.  x2-2xy  4-y2andic24-2icy4-y2. 

14.  x  —  y  4-  13  34  and  »  —  y  —  13  z4. 

15.  x2-5y2-  7z92Uidx2-5y2  +  7zz. 

112.    Square  of  any  Polynomial.     If  we  put  x  for  a,  and 

y  4-  «  f or  b,  in  the  identity 

(a  4-  5)2  =  a2  +  2  a&  4-  #*, 
we  have 

ja  +  (y  4-  *)?2  =  x2  4-  2a  (y  4-  *)  +  (y  4-  *)2, 
or      (x  +  y  +  z)2      =x2  +  2xy  +  2xz  +  y2  +  2yz  +  z2 
=  x2  +  y2  +  z2  +  2xy  +  2xz  +  2yz. 

The  complete  product  consists  of  the  sum  of  the  squares 
of  the  terms  of  the  given  expression  and  twice  the  product 
of  each  term  into  all  the  terms  that  follow  it 


78  SPECIAL  RULES   OF  MULTIPLICATION. 

Again,  if  we  put  a  —  b  for  a,  and  c  —  d  for  b,  in  the  same 
identity,  we  have 

=  (a  -  b)2  +  2  (a  -  &)  (0  -  d)  +  (c  -  d)* 
=  (a2-2ab  +  b2)  +  2a(c-d)-2b(c-d)  +  (c2-2cd  +  d2) 
±=a2-2ab  +  b2  +  2ac-2ad-2bc  +  2bd''rc2-  2cd  +  d2 
=  a2  +  b2  +  c2  +  d2-2ab  +  2ac-2ad-2bc  +  2bd-  2cd. 

Here  the  same  law  holds  as  before,  the  sign  of  each 
double  product  being  +  or  — ,  according  as  the  factors  com- 
posing it  have  like  or  unlike  signs.  The  same  is  true  for 
any  polynomial.     Hence  we  have  the  following  rule : 

Rule  4.  The  square  of  a  polynomial  is  the  sum  of  the 
squares  of  the  several  terms  and  twice  the  product  obtained 
by  multiplying  each  term  into  all  the  terms  that  follow  it. 

Exercise  30. 
Write  the  square  of : 

1.  2x  —  3y.  12.  a -2b -3c. 

2.  a  +  b  +  c.  13.  3a-  b  +  2c. 

3.  x  +  y  —  z.  14.  x  +  2y  —  3z, 

4.  x  —  y  +  z.  15.  x2-y2-\-z\ 

5.  x  +  y  +  5.  16.  x-2y-3z. 

6.  x  +  2y  +  3.  17.  2z-y  +  x. 

7.  a-b  +  c.  18.  s-hy  +  s  +  l. 

8.  3x  -2y  +  4.  19.  x  -  y  +  z  -  1. 

9.  2x-3y  +  4:Z.  20.  4x  +  y  +  z-2. 

10.  x2  +  y2  +  z\  21.    2x-y-z-3. 

11.  2x  —  y-z.  22.   05-2^-3*4-4. 


SPECIAL  RULES   OF  MULTIPLICATION.  79 

113.  Product  of  Two  Binomials.  The  product  of  two  bino- 
mials which  have  the  form  x  +  a,  x  +  b,  should  be  carefully 
noticed  and  remembered. 

1.  (x  +  5)  (x  +  3)  =  x  (x  +  3)  +  5  (x  +  3) 

=  z2  +  3;c  +  5:z  +  15 
=  x2  +  8  a +  15. 

2.  (x  -  5)  (x  -  3)  =  x  (x  -  3)  -  5  (x  -  3) 

=  «2-3£c-  5z  +  15 
=  x2  -8x  +  15. 

3.  (a  +  5)  (cc  -  3)  =  x  (x  -  3)  +  5  (x  -  3) 

=  ar2-3z  +  5a;-15 
=  ic2  + 2^-15. 

4.  (x  -  5)  (a;  +  3)  =  x  (x  +  3)  -  5  (a  +  3) 

=  a2  +  3a-  5* -15 
=  a2  -2x  -15. 

Each  of  these  results  has  three  terms. 

The  first  term  of  each  result  is  the  product  of  the  first 
terms  of  the  binomials. 

The  last  term  of  each  result  is  the  product  of  the  second 
terms  of  the  binomials. 

The  middle  term  of  each  result  has  for  the  coefficient  of  x 
the  algebraic  sum  of  the  second  terms  of  the  binomials. 

The  intermediate  step  given  above  may  be  omitted,  and 
the  products  written  at  once  by  inspection.     Thus, 

1.    Multiply  x  +  8  by  x  +  7. 

8  +  7  =  15 ;  8  X  7  =  56. 
.'.(x  +  8)  (x  +  7)  -  x2  +  15z  +  56. 


80  SPECIAL  RULES   OF  MULTIPLICATION. 

2.  Multiply  x  —  8  by  x  —  7. 

(-8)  +  (-7)  =  -15;    (-8)  (-7)  =  +  56. 
.-.  (x  -  8)  (x  -  7)  =  x2  -  15a  +  56. 

3.  Multiply  x  —  7ybyx  +  6y. 

-7y  +  6y=-y;    (-  7  y)  (6  y)  =  -  42  y\ 
.'.  (x  -  ly)  (x  +  6y)  =  x2  -  xy  -  42y2. 

4.  Multiply  x2  +  6  (a  +  b)  by  a2  -  5  (a  +  J). 

-  5  (a .  +  J)  +  6  (a  +  b)  =  (a  +  b)  ; 

-  5  (a  +  J)  X  6  (a  +  6)  =  -  30  (a  +  J)2. 
.•.^2+6(a  +  J)^x2-5(a  +  J)5  =  a;4+(a+^2-30(a+J)J 


Exercise  31. 

Find  by  inspection  the 

product 

;of: 

1.    (x  +  8)  (x  +  3). 

15. 

(x  +  6y)(x-5y). 

2.    (a  +  8)(«-  3). 

16. 

(x2  -  9)  (x2  +  8). 

3.    (a-7>(«  +  10). 

17. 

(a2  +  2y2)(^-3y2). 

4.    (x  -9)(x-  5). 

18. 

(x2+82/2)(aj2-4y2). 

5.    (a  -  10)  (x  +  9). 

19. 

(ab  -  8)  (a&  +  5). 

6.    (a  -10)  (a  -5). 

20. 

(a£  —  7  icy)  (ab  4-  3  ay). 

7.    (a  -  3a)  (a  +  2a). 

21. 

(z-32/)(z-3y). 

8.    (a +  2  J)  (a-4&). 

22. 

(a  +  6)  (a  +  6). 

9.    (a -12)  (a -3). 

23. 

(a-3£)(a-3£). 

10.    (a +  26)  (a +  46). 

24. 

(a;  —  c)  (a  —  d). 

11.    (a-3o)(a  +  76). 

25. 

(a  +  a)  (x  —  5). 

12.    (a +  25)  (a -9  J). 

26. 

(a:  —  a)  (a  +  b). 

13.    (s-3a)  (s-4a). 

27. 

\(a  +  b)  +  2\\(a  +  b)-± 

14.    (s  +  4*)(«-2«).  28.    \(x+y)-2\\(x+y)+±\. 


SPECIAL  BULES   OF  MULTIPLICATION.  81 

114.    In  like  manner  the  product  of  any  two  binomials 
may  be  written. 

1.  Multiply  2a-bby  Sa  +  Ab. 

(2a-b)(3a  +  4:b)  =  6a2  +  8ab-3ab-4tb* 
=  6a2  +  5ab-U2. 

2.  Multiply  2x  +  3y  by  3x-2y. 

The  middle  term  is 

2x  X  ( —  2y)  +  3y  X  3x  =  5xy. 
.-.  (2  a;  +  3y)  (3x  -2y)  =  6x2  +  5xy-  6y*. 

Exercise  32. 
Find  the  product  of : 

1.  3x  —  y  and  2x  -\-y.  6.  10  a;  —  3y  and  10  a;  —  1  y. 

2.  ±x-3y  tm&3x-2y.  7.  3a2-  2b2  and  2a?  +  3b\ 

3.  5  x  —  4  y  and  3  x  —  4  y.  8.  a2  -\-b2  and  a  —  b.  - 

4.  a; -7?/ and  2a; -5y.  9.  3a2-  262  and  2a  +  3 J. 

5.  11  x  —  2  y  and  7  x  -\-  y.  10.  a2  —  b2  and  a -{- b. 

11.  4a;-fl  and  3a  —  2. 

12.  3  a  —  5  and  x  +  1. 

13.  3 a2  +  a;2  and  4 a2 -a;2. 

14.  2x  +  y  and  x  +  2y. 

15.  3  J +  a  and  2o-3«. 

16.  2  a -{-5  b  and  4  a  —  36. 

17.  4a;  +  3^  and  2a;  -  7y. 

18.  2y  +  3z  emd  3y  —  z. 

19.  2  a;  +  7  y  and  3  a;  —  y. 

20.  3 a  —  2 c  and  2a  —  5  c, 


82  SPECIAL  RULES   OF  DIVISION. 

Division. 

115.  The  following  rule  for  finding  any  required  root  of 
a  monomial  will  be  found  useful  in  solving  examples  in 
division : 

Find  the  required  root  of  the  numerical  coefficient,  and 
divide  the  exponent  of  each  letter  by  the  index  of  the  re- 
quired root. 

Thus,  the  square  root  of  25  »2y4  is  5  xy\ 

116.  Difference  of  Two  Squares. 

a2  —  b2  a2  —  b2 

— -  =  a  —  b:   —  =  a  -\-b.    Hence, 

a  +  b  a  —  b 

Kule  1.  The  difference  of  the  squares  of  two  numbers  is 
divisible  by  the  sum  of  the  numbers,  and  the  quotient  is  the 
difference  of  the  numbers. 

The  difference  of  the  squares  of  two  numbers  is  divisible 
by  the  difference  of  the  numbers,  and  the  quotient  is  the 
sum  of  the  numbers. 

Exercise  33. 
Write  by  inspection  the  quotient  of : 

a2- A 

7. 


2.    — 8. 


a  — 

2 

9-; 

X2 

3  + 

X 

16- 

-a2 

4  + 

a 

x2- 

25 

x  — 

5 

36- 

X2 

6  + 

X 

9a2 

-b2 

4.    — ••  10. 

x  —  o 


6      — — — — —  •  12 

3a -b  '     a-(b-\-c) 


9a4- 

-252/* 

3  a2 

+  5y2 

4z16 

-9y« 

2  xs 

-3/ 

4x10 

-a8 

2xb 

-a4 

flW 

t-x™ 

ab*& 

l  +  z6 

x4a» 

~bw 

x2a* 

-b* 

a2- 

(b  +  c)2 

SPECIAL  RULES   OF  DIVISION.  83 

117.  Sum  and  Difference  of  Two  Cubes.  By  performing 
the  division,  we  find  that 

a8  +  b8  a8  —  b8 

— —  =  a2  —  ab  +  b2 ;  —  =  a2  -t-  ab  +  b2.  Hence, 

a  -f-  b  a  —  b 

Rule  2.  The  sum  of  the  cubes  of  two  numbers  is  divisible 
by  the  sum  of  the  numbers,  and  the  quotient  is  the  sum  of 
the  squares  of  the  numbers  minus  their  product. 

Rule  3.  The  difference  of  the  cubes  of  two  numbers  is 
divisible  by  the  difference  of  the  numbers,  and  the  quotient 
is  the  sum  of  the  squares  of  the  numbers  plus  their  product. 


Exercise  34. 

Write  by  inspection  the  quotient  of : 

«     l-8z8                                       a8b8-c* 
1.    z £ 9.    — -7 

ab  —  c 

a8b8  +  c* 

ab  +  c 
64-f-y8 

4  +  y  ' 
343 -8  a8 

7-2a 
8  a8  +  b« 

2  a +b2 
x«  +  729y* 

x2  +  9y 
a9 -27  b8 

a2 -3b' 

8a8 -64V* 
l  fi  -• 

l  +  3«  2x-±y2 


l-2x 

l  +  8z8 

l  +  2x 

27  a8  -  b8 

3a-b 

27  a8  +  b8 

3a  +  b 

64  x8  +  27  y8 

4x  +  3y 

64  x8-  27  y8 

4:X  —  3y 

1-27  z8 

l-3z 

1  +  27  z8 

2-  i+jt  10- 

6a  —  b 

«•  *££ 

„     64x8  +  27v8 

64 a8- 27 y8 
6.    — ; ^--  14. 

15. 


a 

-b 

a4 

-b4 

a 

+  b 

a5 

-b5 

a 

-b 

a" 

+  &5 

84  SPECIAL  RULES    OF  DIVISION. 

118.    Sum  and  Difference  of  any  Two  Like  Powers.      By 

performing  the  division,  we  find  that 

=  a*-a2b  +  ab2-  £8; 

=  a4  +  asb  -h  a*b2  +  ab*  +  b4\ 

=  a4  -  a8&  +  a2&2  -  a&8  +  £4. 
<z  +  b 

We  find  by  trial  that 

a2  +  J2,  a4  4-  b4,  a6  +  &6,  and  so  on 
are  wotf  divisible  by  a  -{-  b  ov  by  a  —  b.     Hence, 

When  n  is  a  positive  integer,  it  is  proved  in  chap,  vii, 

1.  an  +  bn  is  divisible  by  a  +  b  if  n  is  odd,  and  by  neither 
a  +  b  nor  a  —  b  if  n  is  even. 

2.  an  —  bn  is  divisible  by  a  —  b  if  n  is  odd,  and  by  both 
a  +  b  and  a  —  b  if  n  is  even. 

Note.  It  is  important  to  notice  in  the  above  examples  that  the 
terms  of  the  quotient  are  all  positive  when  the  divisor  is  a  —  b,  and 
alternately  positive  and  negative  when  the  divisor  is  a  +  b  ;  also,  that 
the  quotient  is  homogeneous,  the  exponent  of  a  decreasing  and  of  b 
increasing  by  1  for  each  successive  term. 

Exercise  35. 
Find  the  quotient  of : 

x«-y*  x4-l  m    a;5 +  32 

4. 


x  —  y 


X 

-hi 

X4 

-16 

X 

-2 

a* 

-32 

^    se6  —  V5 
2.  y  • 

x  +  y 

3-    x-1  6'     x-2  9' 


X 

1 

+  2 
—  m4 

1  —  m 

1  +  m6 

CHAPTER  VIL 
FACTORS. 

119.  Rational  Expressions.  An  expression  is  rational 
if  none  of  its  terms  contain  square  or  other  roots. 

120.  Factors  of  Rational  and  Integral  Expressions.  By  fac- 
tors of  a  given  integral  number  in  Arithmetic  we  mean 
integral  numbers  that  will  exactly  divide  the  given  number. 

Likewise,  by  factors  of  a  rational  and  integral  expression 
in  Algebra  we  mean  rational  and  integral  expressions  that 
will  exactly  divide  the  given  expression. 

121.  Factors  of  Monomials.  The  factors  of  a  monomial 
may  be  found  by  inspection.  Thus,  the  factors  of  14  a2b 
are  7,  2,  a,  a,  and  b. 

122.  Factors  of  Polynomials.  The  form  of  a  polynomial 
that  can  be  resolved  into  factors  often  suggests  the  process 
of  finding  the  factors. 

123.  When  the  terms  have  a  common  monomial  factor. 
Kesolve  into  factors  2x2  +  6  xy. 

Since  2  and  x  are  factors  of  each  term,  we  have 

2x2  +  6xy      2x2  ,  6xy 

2x        =Jx-  +  ^  =  X  +  Sy' 
.-.  2x2  +  Qxy  =  2x(x  +  Sy). 

Hence,  the  required  factors  are  2  x  and  x  +  3  y. 


86  FACTORS. 

Exercise  36. 

Resolve  into  factors : 

•1.  3a2 -6a8.  13.  8 a2x2-4a2b  +  12 ahj\ 

2.  2  a2  -4  a.  14.  8  a8b2c2  -  4  aW  +  2  aW. 

3.  5ab-5a2b*.  15.  15  a8*  -  10  aty  +  5  a8*. 

4.  3a2£-4a&2.  16.  a8cy8  +  2  aVy2  -  a2c2/4. 

5.  SxY  +  4,xy.  17.  3&8c8  +  2&2c2-6fo8. 

6.  3a8-a2  +  a.  18.  6a2b  -  Sab  -  12 ab2. 

7.  x*  +  x2y-xy\  19.  5a¥  +  3a2c  +  4aV. 

8.  a*-a*b  +  a2b2.  20.  6  a;8/  +  3  x2y2  -  15  xy\ 

9.  3  a4 -9  a2 -6  a8.  21.  7  aW  -  14  a£c  +  7  ab2c\ 

10.  a£2-6c2  +  fa;.  22.  8x2y2  +  16xyz-24:X2y2»*. 

11.  8  a2b  -  6  a8  -f  4  a&.  23.  a£2c8  -  2  a2fo  +  3  a868c2. 

12.  4  x2y  —  8  a??/2  —  4  icy.  24.  x2y2z2  —  x8y2zz  -f  ic2i/8«. 

124.   When  the  terms  can  be  grouped  so  as  to  show  a  common 
compound  factor. 

1.  Resolve  into  factors  ac  +  ad  +  be  -f  #<£ 

ac  +  ad  +  6c  +  bd  =  (ac  +  ad)  +  (be  +  M)  (1) 

=  a(c  +  d)  +  b(c  +  d)  (2) 

=  (a  +  6)  (c  +  d).  (3) 

Since  one  factor  is  seen  in  (2)  to  be  c .+  d,  dividing  by  c  +  d  we 
obtain  the  other  factor,  a  +  b. 

2.  Resolve  into  factors  3x2  +  6ax  +  bx  +  2ab. 

3&  +  6ax  +  bx  +  2ab  =  (Sx*  +  6ax)  +  (bx  +  2db) 
=  3x  (x  +  2  a)  +  b  (x  +  2a) 
=  (3x  +  6)(x  +  2a). 


FACTORS.  87 

3.  Find  the  factors  of  ac  +  ad  —  be  —  bd. 

ac  +  ad  —  bc  —  bd  =  (ac  +  ad)  —  (bc  +  bd) 
=  a{c  +  d)-b(c  +  d) 
=  (a-b)(c  +  d). 

Note.     Here  the  last  two  terms,  —  be  —  bd,  being  put  within  a 
parenthesis  preceded  by  the  sign  — ,  have  their  signs  changed  to  +. 

4.  Eesolve  into  factors  3  #8  —  5  x2  —  6  x  +  10. 

3x«  -  5x2  -  6x  +  10  =  (3x8  -  5x2)  -  (6x  -  10) 
=  x2(3x-5)-2(3x-5) 
=  (x2-2)(3x-5). 

5.  Eesolve  into  factors  5  xz  —  15  ax2,  —  x  +  3  a. 

6x8  -  15 ox2  -  x  +  3 a  =  (5x3  -  15 ax2)  -  (x  —  3 a) 
=  5x2  (x  —  3  a)  —  1  (x  —  3  a) 
=  (5x2-l)(x-3a). 

6.  Eesolve  into  factors  6  y  —  27  x2y  —  10  x  +  45  a;8. 

6 y  -  27 xfy  —  lOx  +  45 x8  =  6y  —  lOx  —  27 x2y  +  45 x8 

=  (6y  -  lOx)  -  (27 x2y  -  45 x8) 
=  2  (3y  —  5x)  —  9x2(3y  —  5x) 
=  (2-9x2)(3y-5x). 


Exercise  37. 


Eesolve  into  factors : 

1.  ax  —  focVf  ay  —  by. 

2.  ax  —  bx  —  ay  r  by. 

3.  ax  —  cy  —  ay  -\-  ex. 

4.  se2  +  aa?  —  &»  —  ab. 

5.  cc2  +  jcy  —  ax  —  ay. 

6.  x2  —  xy  —  6x  +  6y. 


7.  2a£-3ac-2&y  4-3cy. 

8.  2  x2  —  3  ccy  +  4  aa?  —  6  ay. 

9.  ab-3bc-2ae+6e2. 

10.  xz  -\-  x  —  x2z  —  «. 

11.  a:8  4- 4a:2  4- 3a  4- 12. 

12.  3ae  —  3ax  —  c  +  x. 


88  FACTORS. 

13.  a%  —  abx  —  ac  +-  ex.  23.  a%x  +-  #2c#  —  a2cy  —  fofy 

14.  2a8-3a2-4a;+-6.  24.  3z2  -  5y2  -  6xB  +  10xy2 

15.  aa4  +  bx*  -  aa  -  b.  25.  8  ace  -  10  foe  -  12  a  +  156. 

16.  ax2  +  a2x  +  a  +  x.  26.  6  x4  +  8  x*  -  9  ic2  -  12  x. 

17.  (x-yy  +  2y(x-y).  27.  3c#4-  2aV -  9cx2  +-  6dx. 

18.  H-15cc4-5cc-3ic8.  28.  (a  +  b)(c  +  d)-3c(a  +  b). 

19.  a2  -  «8  +  1  —  x.  29.  bsx  +  bc2x  —  b2cy  -  czy. 

20.  £c8-5»2-f  2a;- 10.  30.  2ac-bc  +  4:a2-2ab. 

21.  a8+-7x2+-3a+-21.  31.  A  +  ac2-«w-ck 

22.  24-«8-s  +  l.  32.  1+-  c  -  c*xy  -  csxy. 

125.  When  a  trinomial  is  a  perfect  square.  A  trinomial  is 
a  perfect  square  if  its  first  and  last  terms  are  perfect 
squares  and  positive,  and  its  middle  term  is  twice  the 
product  of  the  square  roots  of  the  first  and  last  terms. 

Thus,  16  a2  —  24  ab  +  9  b2  is  a  perfect  square. 

The  rule  for  extracting  the  square  root  of  a  perfect 
trinomial  square  is  as  follows : 

Extract  the  square  roots  of  the  first  and  last  terms,  and 
connect  these  square  roots  by  the  sign  of  the  middle  term. 

Thus,  if  we  wish  to  find  the  square  root  of 

16a2-24o6  +  962, 

we  take  the  square  roots  of  16  a2  and  9  b2,  which  are  4  a  and  36, 
respectively,  and  connect  these  square  roots  by  the  sign  of  the  middle 
term.     The  square  root  is  therefore 

4a-S6. 
In  like  manner,  the  square  root  of 

16a2 +  24  06  +  962  is 
4a  +  36. 


FACTORS.  89 


1.  Resolve  into  factors  x2  +  2  xy  +  y2. 

The  factors  of  x2  +  2  xy  +  y2  are 

(x  +  y)  (x  +  y). 

2.  Resolve  into  factors  a;4  —  2  x2y  +  y2. 

The  factors  of  x4  —  2  x2y  +  y2  are 

(x2  -y)  (x2  -  y). 


Exercise  38. 
Resolve  into  factors : 

1.  a2-6ab  +  9b2.  18.  121  a2  +  198 ay  +  81  */2. 

2.  4a2  +  4a&  +  £2.  19.  aW  -  2  a&W5  +  a;16. 

3.  a2-4a&  +  462.  20.  49  -  140  k*  +  100  A;4. 

4.  x2  +  6xy  +  9y2.  21.  49 a2  +  42 ac2  +  9 c4. 

5.  4a2-12az  +  9a2.  22.  81c2-90c  +  25. 

6.  a2  -  10 ab  + 25b2.  23.  121  +  110 a;  +  25 a2. 

7.  4aa-4a  +  l.  24.  144  +  168  3  + 49  z2. 

8.  49  2/2  —  14  yz  +  z2.  25.  36  a2  -  60  ccy  +  25  y2. 

9.  x2  -16  a; +  64.  26.  y2  -  50  yz  +  625  «2. 

10.  9  a:2 +  24  a:?/ +  16  y2.  27.  a;6  -  34  x*  +  289. 

11.  16  a2  +  8  ax  +  x2.  28.  49  a;2  -  112  xy  +  64  y\ 

12.  25 +  80  a; +  64  a;2.  29.  49  a2b2  -  28  abc  +  4  c2 

13.  49  a;2 -28^2/  + 4  2/2.  30.  121  a;2  -  286  xy  +  169  y\ 

14.  1-20  6  +  100&2.  31.  4a4  +  20a2a;2  +  25a;4. 

15.  81a2  +  126a&  +  49  62.  32.  (x  +  y)2-4z(x  +  y)  +  4:Z2. 

16.  m2^2  -  IQmna2  +  64a4.  33.  (a  -  b)2  -  6  (a  -  b)  +  9. 

17.  4a2-20aaj  +  25a;2.  34.  (a  +  c)2  +  10(a  +  c)  +25. 


90  FACTORS. 

126.   When  a  binomial  is  the  difference  of  two  squares. 

The  difference  of  two  squares  is  the  product  of  two 
factors,  which  may  be  found  as  follows : 

Take  the  square  root  of  the  first  term  and  the  square  root 
of  the  second  term. 

The  sum  of  these  roots  will  form  the  first  factor. 

The  difference  of  these  roots  will  form  the  second  factor. 

Resolve  into  two  factors  16  x2  —  9  y*. 

The  square  root  of  16  x2  is  4  x. 
The  square  root  of  9  y6  is  3  2/3. 
The  sum  of  these  roots  is  4  x  +  3  y8. 
The  difference  of  these  roots  is  4  x  —  3  y8. 
Therefore,  16  x2  —  9  y6  =  (4  x  +  3  ys)  (4  z  —  3  y8). 

Exercise  39. 
Eesolve  into  factors : 

1.  a2-4.    '  13.    25-16a2-  25.  8lx2-±y\ 

2.  1-x2.  14.    16-25y2.  26.  64  a4  -  b\ 

3.  x2-9y2.  15.    a2b2-l.  27.  m2n2  -  36. 

4.  4a2-4962.        16.    x2  -  100.  28.  z4  -  144. 

5.  a2 -4  ?/2.  17.    121  a2-  36  62.  29.  z2-25. 

6.  49-  100  y2.       18.    49  a14-?/12.  30.  25-64y2. 

7.  l-49ic6.  19.    64,  a2 -9  b6.  31.  16x17-9xy*. 

8.  4-121?/8.         20.    81a464-c4.  32.  25x10-16a*x\ 

9.  1-169  a6.         21.    4a2c-9c5.  33.  36  a2x2  -  49  a4. 

10.  a262  -  4  c6.  22.    20  a3b*  -  5  ab.   34.    cc2  -  16  y\ 

11.  9x8-a6.  23.    3a5-12aV.    35.    1-400  a4. 

12.  4a16-2/20.  24.    9  a2 -Sib2.       36.    4a2c-9c8. 


FACTORS.  91 

127.    If  the  squares  are  compound  expressions,  the  same 
method  may  be  employed. 

1.  Eesolve  into  factors  (x  +  3y)2  —  16  a2. 

The  square  root  of  the  first  term  is  x  +  3  y. 

The  square  root  of  the  second  term  is  4  a. 

The  sum  of  these  roots  is  x  +  3  y  +  4  a. 

The  difference  of  these  roots  is  x  +  3  y  —  4  a. 

Therefore,  (x  +  Sy)2  —  16a2  =  (x  +  3y  +  4a)  (x  +  Sy  —  4a). 

2.  Eesolve  into  factors  a?  —  (3b  —  5c)2. 

The  square  roots  of  the  terms  are  a  and  (3  6  —  5  c). 
The  sum  of  these  roots  is  a  +  (3  6  —  5  c),  or  a  +  3  b  —  5  c. 
The  difference  of  these  roots  is  a  —  (3  6  —  5  c),  or  a  —  3  b  +  5  c. 
Therefore,  a2  -  (36  -  5c)2  =  (a  +  36  -  5c)  (a  -  36  +  5c). 

If  the  factors  have  like  terms,  these  terms  should  be 
llected  so  as  to  give  the  results  in  the  simplest  form. 

3.  Eesolve  into  factors  (3  a  +  5  b)2  -  (2  a  -  3  b)2. 

The  square  roots  of  the  terms  are  3  a  +  5  6  and  2  a  —  3  6. 
The  sum  of  these  roots  is  (3  a  +  5  6)  +  (2  a  —  3  6), 

or3a  +  56  +  2a  —  36  =  5a  +  2  6. 
The  difference  of  these  roots  is  (3  a  +  5  6)  —  (2  a  —  3  6), 

or3a  +  56  —  2a  +  36  =  a  +  8  6. 
Therefore,  (3a  +  5  6)2  -  (2  a  -  3  6)2  =  (5  a  +  2  6)  (a  +  8  6). 

Exercise  40. 

Eesolve  into  factors : 
(x  +  y)2-z2.  5.    (a  -  b)2  -  (c  -  d)2. 

(x-y)2-z2.  6.    (2  a  +  b)2  -  25  c2. 

.    (x-2y)2-4z2.  7.    (x  +  2y)2-(2x-y)2. 

(a  +  3b)2  -  16c2.  8.    (x  +  3)2-(3x-  4)2. 


92  FACTORS. 

9.  x2-(y-z)2.  15.  (a  +  b-cy-(a-b-c)\ 

10.  a2 -(3b -2c)2.  16.  (a-3a)2-(3a-2;c)2. 

11.  b2-(2a  +  3c)2.  17.  (2a- l)2- (3a +  1)2. 

12.  l-(a  +  5&)2.       '  18.  (x-5)2-(x  +  y-5)2. 

13.  9a2-  (x  -3c)2.  19.  (2a  +  £-c)2-(a-2&+c)2, 

14.  16a2-(2y-3z)2.  20.  (a  +  2b-3cy-(a  +  5cy. 

128.  By  properly  grouping  the  terms,  compound  expres- 
sions may  often  be  written  as  the  difference  of  two  squares, 
and  the  factors  readily  found. 

1.  Kesolve  into  factors  a2  —  2ab  -\- b2  —  9 c\ 

a2  —  2ab  +  62  —  9c2  =  (a2 -  2ab  +  62)  —  9c2 
=  (a  -  b)2  -  9  c2 
=  (a-b  +  3c)(a-b  —  Sc). 

2.  Kesolve  into  factors  12  ab  +  9  x2  -  4  a2  -  9  £2. 

Here  12  ab  shows  that  it  is  the  middle  term  of  the  expression  which 
•    has  in  its  first  and  last  terms  a2  and  62,  and  the  minus  sign  before  4  a2 
and  9  b2  shows  that  these  terms  must  be  put  in  a  parenthesis  with  the 
minus  sign  before  it,  in  order  that  they  may  be  made  positive. 

Therefore,  the  arrangement  will  be 
9s2-(4a2-12a&  +  9&2)  =  9x2  -  (2a  -  36)2 

=  (3s  +  2a  — 3&)(3»  — 2a  +  36). 

3.  Kesolve  into  factors  -  a2  +  b2  -  c2  +  d?  +  2ac  +  2bd. 

Here  2  bd  must  be  grouped  with  62  and  d2,  and  2  ac  with  —  a4  and 
—  c2}  and  this  last  group  put  in  a  parenthesis  preceded  by  — . 

-  a2  +  b2  -  c2  +  d2  +  2  ac  +  2  bd 

=  (b2  +  2  6d  +  d2)  -  (a2  -  2  ac  +  c2) 

=  (6  +  d)2  -  (a  -  c)2 

sb  (&  +  d  +  «  —  c)  (b  +  4*-  a  +  c). 


FACTORS.  93 

Exercise  41. 
Resolve  into  factors : 

1.  a2  +  2ab  +  b2-4:C2.  5.    a2  -  x2  -  y2  -  2  xy. 

2.  x2-2xy  +  y2-9a2.          6.    l-a2-2ab-b2. 

3.  b2-x2  +  ±ax-±a2.  7.    a2 +  b2 +  2ab  -  l§a2b\ 

4.  4a2  +  4a&  +  &2-#2.  8.    Ax2  -  9  a2  +  6a  -  1. 

9.  a2  +  &2-c2-d2-2a&-2ctf\ 

10.  x2  +  y2  -  2xy  -  2ab  -  a2  -  b2. 

11.  9a2-6a;  +  l-a2-4a&-462 

12.  a2  +  2ab-x2-6xy-9y2  +  b2. 

13.  a2-  2#  +  l-&2  +  2fy  -y2. 

14.  9-6a-ha2-a2-8a&-16&2. 

15.  4-4«  +  a2-  ±ab-b2-±a\ 

16.  a4  -  a2  -  9  +  V  +  6  a  -  2  a262. 

17.  4a2  +  9c2-12ac  +  12^-9&2-4d2 

18.  4cc2-42/2-4a;  +  l-«2  +  42/«. 
19..  ±xy-x2  +  l  -±y2. 

20.  c2-2ac-l  +  2£-}-a2-&2. 

21.  4ac-l-6a;-9a;2  +  a2  +  4c2. 

22.  4-9ic2-42/2  +  12icy. 

23.  4a4-12a2-9z2  +  12  2/z-4?/2  +  9. 

24.  a4-  b2-4:xi-6a2c  +  4;bx2  +  9c2. 

25.  25a2-l-10a&-9a2a2  +  &2+6«sc. 

26.  16£c4  +  30cc8  +  8c2x2-25x6-9  +  c4. 

27.  4  a4  +  9  b«  -  12  a268  -  81  a2^6. 


94  FACTORS. 

129.    A  trinomial   in   the   form    a4  +  a2b2  +  64    can    be 

written  as  the  difference  of  two  squares. 

Since  a  trinomial  is  a  perfect  square  when  the  middle 
term  is  twice  the  product  of  the  square  roots  of  the  first 
and  last  terms,  it  is  obvious  that  we  must  add  a%2  to  the 
middle  term  of  a4  +  a2b2  +  &4  to  make  it  a  perfect  square. 
We  must  also  subtract  a2b2  to  keep  the  value  of  the 
expression  unchanged. 

1.  Eesolve  into  factors  a4  4-  a2b2  -f-  b\ 

a4  +  a262  +  &4  =  a4  +  2  a262  +  64  -  a262 
=  (a2  +  62)2  -  a262 
=  (a2  +  62  +  ab)  (a2  +  62  -  ab) 
=  (a2  +  a6  +  62)  (a2  -  ab  +  52). 

2.  Resolve  into  factors  4  a4  -  37  »y  -f-  9  y4. 

Twice  the  product  of  the  square  roots  of  4  x4  and  9  y*  is  12  x*y*., 
We  may  separate  the  term  —  'Sixty2  into  two  terms,  —  12x2y2  and 
—  25  x2y2,  and  write  the  expression 

(4  x4  -  12  x2^2  +  9  y4)  -  25  x2y2 

=  (2x2  — 32/2)2-25xV 

=  (2x2  —  Sy2  +  6xy)  (2x2  —  Sy2  —  5xy) 

=  (2x2  +  hxy  —  Sy2)  (2x2  —  hxy  —■  Sy*). 

Exercise  42. 
Resolve  into  factors 

1.  x*  +  xy  +  y*.  6.  9a*  +  26a2b*  +  25b\ 

2.  »4  +  a2  +  1.  7.  4x*  -  21x2f +  9y\ 

3.  9a4-15a2  +  l.  8.  4a4-29a2c2  +  25c4. 

4.  16a4  -  17a2 +  1.  9.  4 a*  +  16 a2c2  +  25 c4. 

5.  4a4-13a2  +  l.  10.  25 a4  +  31  *y  +  16 y* 


FACTOBS.  95 

130.   When  a  trinomial  has  the  form  x2  +  ax  +  6. 

Where  a  is  the  algebraic  sum  of  two  numbers,  and  is 
either  positive  or  negative ;  and  b  is  the  product  of  these 
two  numbers,  and  is  either  positive  or  negative. 

Since  (x  +  5)  (x  +  3)  =  x2  +  Sx  +  15, 

the  factors  of    x2  -f  8  a;  +  15  are  a;  +  5  and  sc  +  3. 

Since  (x  +  5)  (a  -  3)  =  x2  +  2x  -  15, 

the  factors  of    x2  +  2  a;  —  15  are  cc  ■+■  5  and  a;  —  3. 

Hence,  if  a  trinomial  of  the  form  x2  -\-  ax  +  b  is  such  an 
expression  that  it  can  be  resolved  into  two  binomial  factors, 
the  first  term  of  each  factor  will  be  x ;  and  the  second 
terms  of  the  factors  will  be  two  numbers  whose  product  is  b, 
the  last  term  of  the  trinomial,  and  whose  algebraic  sum  is  a, 
the  coefficient  of  x  in  the  middle  term  of  the  trinomial. 

1.  Eesolve  into  factors  x2  +  11  x  -J-  30. 

We  are  required  to  find  two  numbers  whose  product  is  30  and 
whose  sum  is  11. 

Two  numbers  whose  product  is  30  are  1  and  30,  2  and  15,  3  and  10, 
6  and  6  ;  and  the  sum  of  the  last  two  numbers  is  11.     Hence.; 

x2  +  11  x  +  30  =  (x  +  5>(a  +  6). 

2.  Eesolve  into  factors  x2  —  7  x  +  12. 

We  are  required  to  find  two  numbers  whose  product  is  12  and 
whose  algebraic  sum  is  —  7. 

Since  the  product  is  +  12,  the  two  numbers  are  both  positive  or  both 
negative ;  and  since  their  sum  is  —  7,  they  must  both  be  negative. 

Two  negative  numbers  whose  product  is  12  are  —  12  and  —  1,-6 
and  —  2,-4  and  —  3  ;  and  the  sum  of  the  last  two  numbers  is  —  7. 
Hence, 

x2  -  Ix  4- 12  =  (x-  4)  (x  -  3). 


96  FACTORS. 

3.  Resolve  into  factors  x2  +  2  x  ~  24* 

We  are  required  to  find  two  numbers  whose  product  is  —  24  and 
whose  algebraic  sum  is  2. 

Since  the  product  is  —  24,  one  of  the  numbers  is  positive  and  the 
other  negative ;  and  since  their  sum  is  +2,  the  larger  number  is 
positive. 

Two  numbers  whose  product  is  —  24,  and  the  larger  number  posi- 
tive, are  24  and  —  1,  12  and  —  2,  8  and  —  3,  6  and  —  4 ;  and  the  sum 
of  the  last  two  numbers  is  +  2.     Hence, 

x2  -f-  2x  -  24:  =  (x  +  6)  (x  -  4). 

4.  Resolve  into  factors  x2  —  3  x  —  18. 

Since  the  product  is  —  18,  one  of  the  numbers  is  positive  and  the 
other  negative ;  and  since  their  sum  is  —  3,  the  larger  number  is 
negative. 

Two  numbers  whose  product  is  —  18,  and  the  larger  number  nega- 
tive, are  —  18  and  1,  —  9  and  2,-6  and  3 ;  and  the  sum  of  the  last 
two  numbers  is  —  3.     Hence, 

x2  -  Sx  -  18  m  (x  -  6)  (x  +  3). 

5.  Resolve  into  factors  x2  —  10  xy  +  9  y2. 

We  are  required  to  find  two  numbers  whose  product  is  Oy2  and 
whose  algebraic  sum  is  —  10  y. 

Since  the  product  is  +  9y2,  and  the  sum  —  10  y,  the  last  two  terms 
must  both  be  negative. 

Two  negative  numbers  whose  product  is  Qy2  are  —  Qy  and  —  y, 
—  Sy  and  —  Sy ;  and  the  sum  of  the  first  two  numbers  is  —  10  y. 
Hence, 

x2  -  lOxy  +  9y2^(x-  9y)  (x  -  y). 

131.  From  these  examples  it  will  be  seen  that  the 
following  statements  are  true : 

1.  If  the  third  term  of  a  given  trinomial  is  negative,  the 
second  terms  of  its  binomial  factors  have  unlike  signs. 

2.  If  the  third  term  of  a  given  trinomial  is  positive,  the 
second  terms  of  its  binomial  factors  have  the  same  sign,  and 
this  sign  is  the  sign  of  the  middle  term. 


FACTORS.  97 

Exercise  43. 
Resolve  into  factors : 

1.  x2  +  8x  +  15.  24.  x2  +  Tx  +  10. 

2.  x2  -8x  +  15.  25.  ic2-7#  +  10. 

3.  a;2  +  2a; -15.  26.  z2  +  3a;-10. 

4.  x2-3ic-10.  27.  a2  +  ax  -6a2. 
6.  a;2  +  5  ax  +  6  a2.  28.  x2  —  ax  —  6  a2. 

6.  x2- 5  ax +  6  a2.  29.  a2  +  5  scy  -f  4  y2 

7.  £c2-2x-15.  30.  x2- 3x7/ -Ay2. 

8.  x2  +  5«+6.  31.  a2-5ajy  +  4y2. 

9.  x2-5x  +  6.  32.  x2  +  3xy-±y\ 

10.  £C2  +  x-6.  33.  cc2  +  3a;y  +  2y2. 

11.  x2  -  x  —  6.  34.  a2-7ab  +  10b2. 

12.  a;2  +  6a;  +  5.  35.  a2x2  -  3 aa;  -  54. 

13.  a;2 -6a; +  5.  36.  a;2 -7a; -44. 

14.  a;2  +  4a;-5.  37.  x2  +  x-132. 

15.  a;2 -4a; -5.  38.  x2  — 15a; +  50. 

16.  a;2  +  9a; +  18.  39.  a2 -23  a +  120. 

17.  a;2 -9a; +  18.  40.  a2 +  17  a -390. 

18.  a;2  +  3a; -18.  41.  c2  +  25c-150. 

19.  a;2 -3a; -18.  42.  g2-  58c  +  57. 

20.  x2  +  9x  +  8.  43.  a4-lla2£8  +  30&». 

21.  x2-9x  +  8.  44.  «2  +  9«?/  +  202/2. 

22.  x2  +  Tx-8.  45.  £cV  +  19a;^+48«* 

23.  a;2 -7  a; -8.  46.  a2b2  —  13  abc  +  22  c\ 


98  FACTORS. 

132.    When  a  trinomial  has  the  form  ax2  +  bx  +  c. 

1.  Find  the  factors  of  8  x2  -  22  x  -  21. 

Multiply  by  8,  the  coefficient  of  x2,  and  write  the  result  in  the 
following  form: 

(8x)2-22  X  8x  — 168. 
Put  z  for  8  x,  and  we  have 

z2  -  22  z  -  168. 
Resolve  this  expression  into  two  binomial  factors  (§  130) 
(2  -  28)  (s  +  6). 

Since  we  have  multiplied  by  8,  and  put  z  for  8  x,  we  must  reverse 
this  process.     Hence,  put  8  x  for  z  and  divide  by  8,  and  we  have 
(8s  —  28)  (8s +  6), 
8 

As  4  is  a  factor  of  (8x  —  28),  and  2  is  a  factor  of  (8x  +  6),  we 
divide  by  8  by  dividing  the  first  factor  by  4  and  the  second  factor  by 
2,  thus 

(8  x  —  28)  (8  s  +  6) 

4X2 
=  (2x-7)(4x  +  3). 

2.  Find  the  factors  of  24  a;2  -  70  xy  -  75  y\ 

Multiply  by  24,  (24  x)2  —  70  y  X  24  x  —  1800  y2. 

Put  z  for  24  x,  z2  —  7Qyz  —  1800  y2. 

Resolve  into  factors,  (z  —  90  y)  (z  +  20  y).  (§  130) 

Put  24  x  for  z,  (24  x  —  90  y)  (24  x  +  20  y). 

Divide  by  6X4,        (4x  —  16y)  (6x  +  6y). 

3.  Find,  the  factors  of  12  cc2  -  23  a^  -f-  10  y2. 

Multiply  by  12,  (12  x)2  -  23  y  X  12  x  +  120  y2. 

Put  z  for  12  x,  z2  —  2Syz  +  120  y2. 

Resolve  into  factors,  (z  —  15  y)  (z  —  8  y).  (§  130) 

Put  12  x  for  z,  (12  x  —  15y)(12x  —  $y). 

Divide  by  3  X  4,         (4*  —  5  y)  (3  x  —  2  y). 


FACTORS.  99 

Exercise  44. 
Eesolve  into  factors  : 

1.  2x2  +  5x  +  3.  25.  2x2  +  5xy  +  2y2. 

2.  3x2-x~2.  26.  6x2-7bx-3b2. 

3.  5x2-Sx  +  S.  27.  8a2-f  14a&-15&2. 

4.  6x2  +  7x  +  2.  28.  6a2-19ac  +  10c2. 

5.  6x2-x-2.  29.  8a2  +  34z?/  +  21y2. 

6.  15z2  +  14z-8.  30.  Sx2-22xy-21y2. 

7.  8x2-10a  +  3.  31.  6x2  +  19  xy -7 y2. 

8.  18«2  +  9cc-2.  32.  11  a2  -  2Sab  +  2£2. 

9.  12cc2-5x-2.  33.  2c2  -  IScd  +  6d2. 

10.  12z2-7cc  +  l.  34.  6y2  +  7yz-3z2. 

11.  12x2-x-l.  35.  15a:2-26a;y  +  8y2 

12.  3x2-2x-5.  36.  9av2  +  6a:y-8?/2. 

13.  3x2  +  ±x-4:.  37.  6z2-a^-35y2. 

14.  6a2  +  5;c-4.  38.  10 a;2- 21^-10^. 

15.  4z2-fl3z  +  3.  39.  Ux2-55xy  +  21y2. 

16.  4ic2  +  ll£c-3.  40.  6x2-23xy  +  20y*. 

17.  4a;2 -4a; -3.  41.  6 x2  +  35 ay  -  6 y2. 

18.  4x2  +  8£C  +  3.  42.  2±x2-14:xy-  5y2. 

19.  6a2x2  +  az-l.  43.  24  a2 -38  ay  +  15  y2. 

20.  6a2  +  17  a +  12.  44.  24a2-2ay-15y2. 

21.  12 x2-  13 a -14.  45.  36x2-19xy-6y2. 

22.  10a2 -23a -5.  46.  15x2  +  19xy  +  6y2. 

23.  8c2 -|- 53c -21.  47.  12 x2  +  31  xy  -  15 y2. 

24.  8«2-37s-15.  48.  5y2  +  13y-6. 


100  FACTORS. 

133.   When  a  binomial  is  the  sum  or  difference  of  two  cubes. 


Since 

— -J-j.  =  «2  _  ab  +  h2 

(§117) 

and 

o?  —  hz        ,  ,      ^  ,  M 

r  =  a2  +  «&  4-  62 ; 

a  —  6 

(§117) 

.-.  a8  +  £8  =  (a  +  &)  (a2  _  ah  +  ja) 

(1) 

and 

a8  _  &8  =  (a  _  j)  (a*  +  a&  +  £2). 

(2) 

Therefore,  the  sum  of  two  perfect  cubes  is  divisible  by 
the  sum  of  their  cube  roots,  and  the  difference  of  two  per- 
fect cubes  is  divisible  by  the  difference  of  their  cube  roots. 

1.  Eesolve  into  factors  8  a8  +  27  b\ 

The  cube  root  of  8  a8  is  2  a,  and  of  27  6s  is  3  6s. 
By  putting  2  a  for  a  and  3  6s  for  6  in  (1),  we  have 
(2a)«  +  (362)3  =  (2a  +  362)  (4a2  -6a62  +  96*). 

2.  Eesolve  into  factors  125  sc8  —  1. 

The  cube  root  of  125  x8  is  5x,  and  of  1  is  1. 
By  putting  5x  for  a  and  1  for  6  in  (2),  we  have 
126 x8  -  1  =  (5x  -  1)  (25 x2  +  5x  +  1). 

3.  Eesolve  into  factors  x*  -f  y9. 

The  cube  root  of  x6  is  x2,  and  of  y9  is  y8. 

By  putting  x2  for  a  and  ys  for  6  in  (1),  we  have 

X«  +  y9  =  (x*  +  y2)  (X4  —  Xty8  +  y«). 

4.  Eesolve  into  factors  (x  —  y)z  +  «8. 

The  cube  root  of  (x  —  y)*  is  x  —  y,  and  of  z8  is  z. 

By  putting  x ;  -  -  y  for  a  and  z  for  6  in  (1),  we  have 

(x  -  yf  +  z»  =  [(x  -  y)  +  z]  [(x  -  y)2  -  (x  -  y)  z  +  z2] 

=  (x  —  y  +  z)  (x2  —  2  xy  +  y2  —  xz  +  yz  +  a:2). 


factors.  [  t ,  ,\\ ; , ,  > ; '  ', ,  •  •  \10l 

Exercise  45. 
Kesolve  into  factors : 

1.  a8  +  8b8.           5.  27x8y8-l.           9.    216  a6 -J8. 

2.  a8 -27  a6.         6.  a*  +  27 b8.            10.    64 a8-  27 b\ 

3.  a8 +  64.            7.  a8y8-64.            11.    343 -a8. 

4.  125  a8 +  1.        8.  64  a6  +  125  b8.     12.    a8b8  +  343. 

13.  8  a8  -  b\  33.  8  x8  -  (x  -  y)\ 

14.  216m8  +  ?i6.  34.  8  (x  +  y)8  +  z\ 

15.  xY  -  512  z3.  35.  (a  -f  6)8  -  (a  -  &)•. 

16.  729  a6  +  216  c6.  36.  (a2  -  3)8  -  z8 

17.  729  2/8- 64  s8.  37.  68  +  (a  -  e)8. 

18.  512  a8 -1.  38.  (z-l)8-(2-l)8. 

19.  (a  +  6)8-l.  39.  (a-Sby-c8. 

20.  (a-6)8  +  l.  40.  (2-3a)8  +  68. 

21.  l-(a-by.  41.  (3  +  5)8-a8. 

22.  27  a6  +  125  y8.  42.  (y  -  z)8  -f  (y  +  z)8. 

23.  216  +  343  y\  43.  (2x  +  y)8-  (x-  y)\ 

24.  a15 -125  a6.  44.  l-(7a-56)8. 

25.  8ic18  +  273/12.  45.  (3a  +  y)8-8«8. 

26.  a16-21669.  46.  (2a-b)8-c8. 

27.  c8d8  -  343  26.  47.  x8  -  (y  -  z)\ 

28.  ^¥  +  8.  48.  125-  (x  — 2  y)8. 

29.  8  ay*9  -  1.  49.  (2x  +  y)8  -  27  z8. 

30.  27a;y^6  +  l.  50.  (a  —  3)8  -  z8. 

31.  64a:8-125?/9.  51.  (7a-6)8  +  c8. 

32.  125 a6 -}- 27 y15.  52.  (3c-2y)8-8z8. 


1,02    •  FACTORS. 


Theory  of  Divisors. 

134.  Theorem.  The  expression  x  —  y  is  an  exact  divisor 
of  xn  —  yn  when  n  is  any  positive  integer. 

Since  —  a"-1  y  +  xn  - 1  y  =  0,  (§34) 

xn  —  yn  =  xn  —  xn~ 1  y  +  xn  ~ 1  y  —  y™. 

Taking  out  xn~ 1  from  the  first  two  terms  of  the  right  side,  and  y  from 
the  last  two  terms,  we  have 

X7l  —  yn  =  j£»-l    (X  _  y)  -{-  y  (rffl-  1  _  y»-l). 

Now  a;  —  y  is  an  exact  divisor  of  the  right  side,  if  it  is  an  exact  divisor 
of  xn  - 1  —  yn  - 1 ;  and  if  x  —  y  is  an  exact  divisor  of  the  right  side,  it 
is  an  exact  divisor  of  the  left  side ;  that  is,  x  —  y  is  an  exact  divisor 
of  xn  —  yn  if  it  is  an  exact  divisor  of  xn  -  J  —  yn  -  K 

But  x  —  y  is  an  exact  divisor  of  xz  —  y*  (§  117),  therefore  it  is  an 
exact  divisor  of  sc4  —  y* ;  and  since  it  is  an  exact  divisor  of  ce4  —  y4,  it 
is  an  exact  divisor  of  x5  —  y5  ;  and  so  on,  indefinitely. 

The  method  employed  in  proving  this  Theorem  is  called 
Proof  by  Mathematical  Induction. 

135.  The  Factor  Theorem.  If  a  rational  and  integral 
expression  in  x  vanishes,  that  is,  becomes  equal  to  0,  when  r 
is  put  for  x,  then  x  —  r  is  an  exact  divisor  of  the  expression, 

Given  axn  +  bxn  - 2  + +  hx  -f  k  (1) 

By  supposition,       ar"  +  brn-i  + +  hr  +  k  =  0.  (2) 

By  subtracting  (2)  from  (1),  the  given  expression  assumes  the  form 

a  (xn  —  r«)  +  b  (x"-1  —  r"-1)  + +  h  (x  —  r). 

But  x  —  r  is  an  exact  divisor  of  xn  —  r»,  xn-1  —  r»  - J,  and  so  on. 
Therefore,  x  —  r  is  an  exact  divisor  of  the  given  expression. 

Note.  If  x  —  r  is  an  exact  divisor  of  the  given  expression,  r  is  an 
exact  divisor  of  k ;  for  fc,  the  last  term  of  the  dividend,  is  equal  to  r, 
the  last  term  of  the  divisor,  multiplied  by  the  last  term  of  the  quotient, 
Therefore,  in  searching  for  numerical  values  of  x  that  will  make  the 
given  expression  vanish,  only  exact  divisors  of  the  last  term  of  the 
expression  need  be  tried. 


FACTORS.    ;  , ,  /,\  >,,;;'  *' , ''  i  '\i08 

1.  Kesolve  into  factors  xs  -f-  3  x2  —  13  x  —  15. 

The  exact  divisors  of  15  are  1,  —  1,  3,  —  3,  5,  —  5,  15,  —  15. 
If  we  put  1  for  x  in  x8  +  3x2  —  13  x  —  15,  the  expression  does  not 
vanish.     If  we  put  —  1  for  x,  the  expression  vanishes. 
Therefore,  x  —  (—  1),  that  is,  x  +  1,  is  a  factor. 
Divide  the  expression  by  x  +  1,  and  we  have 

x3  +  3x2  -  13x  -  15  =  (x  +  1)  (x2  +  2x  -  15) 
=  (x  +  1)  (x  -  3)  (x  +  5). 

Note.     An  expression  can  sometimes  be  resolved  into  three  or 
more  factors. 

2.  Kesolve  into  factors  xs  —  26  x  —  5. 

By  trial  we  find  that  the  only  exact  divisor  of  —  5  that  makes  the 
expression  vanish  is  —  5. 

Therefore,  divide  by  x  +  5,  and  we  have 

x3  -  26x  -  5  =  (x  +  5)  (x2  -  5  x  -  1). 

As  neither  +  1  nor  —  1,  the  exact  divisors  of  —  1,  will  make 
x2  —  5  x  —  1  vanish,  this  expression  cannot  be  resolved  into  factors. 

Exercise  46. 
Eesolve  into  factors : 

1.  a8 -10a -3.  10.  xB-3x2  +  4:X-2. 

2.  x»-26x  +  5.  11.  «8  +  9ic2  +  16ic  +  4. 

3.  a8-15z-4.  12.  xs  +  2x2-S4,x-5. 

4.  z8-8;z  +  3.  13.  4:x*-12x2  +  9x-l. 

5.  z8  +  2x2  +  9.  14.  x*-2x2-23x  +  60. 

6.  x»-3x  +  2.  15.  6ic8-23ic2  +  16cc-3. 

7.  z8-  12 x  +  16.  16.  a;8-10a;2H-33£c-36. 

8.  x8  +  4,x2-5.  17.  x*  +  7 ar2  +  12  a  +  4. 

9.  4a8-7a  +  3.  18.  x8  +  5 a;2  +  7 a  -f  2. 


104  :  <<!  ,  FACTORS. 

136.  A  compound  expression  involving  x  and  y  is  divis- 
ible by  x  —  y  if  the  expression  vanishes  when  +  y  is  put 
for  cc ;  and  is  divisible  by  x  -f-  y  if  the  expression  van- 
ishes when  —  y  is  put  for  x. 

If  n  is  a  positive  integer,  prove  by  the  Factor  Theorem : 

1.  xn  +  yn  is  never  divisible  by  x  —  y. 

Put  y  for  x  in  icn  +  yn ;  then  xn  +  yn  =  yn  +  y7l  =  2yn. 
As  2  y"  is  not  zero,  xn  +  yw  is  not  divisible  by  a;  —  y. 

2.  a;"  —  yn  is  divisible  by  a;  +  y,  if  iz  is  even. 

Put  —  y  for  x  in  ccn  —  yra,  then  xn  —  y71  =  (—  y)n  —  y71. 
If  n  is  even,  (—  y)re  =  yn,  and  (—  y)"  —  yn  —  yn  —  ynm 
As  yn  —  yn  =  0,  xn  —  yn  is  divisible  by  x  +  y,  if  n  is  even. 

3.  xn  +  y"  is  divisible  by  a;  +  y,  if  22  is  odd. 

Put  —  y  for  x  in  «ra  +  y71,  then  xn  +  ?/n  =  (—  y)n  +  y71. 
If  n  is  odd,  (—  y)n  =  —  yn,  and  (—  y)»  +  yn  =  —  yn  +  y*. 
As  —  y"  +  y71  =  0,  zn  +  y71  is  divisible  by  cc  +  y,  if  /7  is  odd. 

From  §  134  and  these  three  cases,  we  have 

1.  For  all  positive  integral  values  of  ny 

xn_yn  =  (x_  y)  ^n-1  +  xn-2y  +  Xn-Zy*  4. 4.  yn-iy 

2.  For  all  positive  even  integral  values  of  n, 

xn  —  yn  =  (x  -f  y)  (a;n_1  —  xn~2y  +  xn~zy2  — —  #*_1). 

3.  For  all  positive  odd  integral  values  of  n, 

x»  +  yn  =  (x  +  y)  (xn~l  —  xn~2y  +  a;B-y  - +  y""1)- 

4.  xn  +  2/n  is  never  divisible  by  x  —  y ;  and  is  not  divisible 
by  x  -^  y,  if  n  is  even. 

Note.  In  applying  the  preceding  rules  for  resolving  an  expression 
into  factors,  if  the  terms  have  a  common  monomial  factor,  this  factor 
should  be  removed  first. 

When  an  expression  can  be  expressed  as  the  difference  of  two  per- 
fect squares,  the  method  of  §  126  should  be  employed  in  preference  to 
any  other. 


FACTORS.  IO16 

Exercise  47.  —  Review. 
Resolve  into  factors : 

1.  a3  -9  a.  23.  x2y2  -  4  xyA  -  3  x2y*. 

2.  ±xA-x2.  24.  x*  +  x2  +  x  +  l. 

25.  Sx2-\-x-2. 

26.  a2 -24  a; +  95. 

27.  9  a2 +  12  a +  4. 

28.  a2-£2-c2  +  2fo. 

29.  x2  —  2y  +  2x-xy. 

30.  3a8  +  2a;2-9a;-6. 

31.  (x-y)2-b2. 

32.  m2-2raw  +  7&2-  1. 

33.  a2  +  2aa;  +  3&z  +  6a&, 

34.  z2  +  ra2-w2-2ma;. 

35.  9xA  +  21x2y2  +  25y\ 

36.  x2  —  4t  +  y2  +  2xy. 

37.  6z2-a;-77. 

38.  x2-8xy-65y\ 

39.  jc4-  7a:2  +  l. 

40.  1  -  a2  -  b2  -  2  ah. 

41.  3a;4-  6x*  +  9x2. 

42.  a:8-5a:2-2;c  +  10. 

43.  x2  -\-  ax  —  bx  —  ab. 

44.  2  a;2  —  3  xy  +  4  aa;  —  6  ay. 


3. 

x12  +  y12. 

4. 

x4  —  y\ 

5. 

x9  +  y\ 

6. 

x6  —  y\ 

7. 

36 a2- 49 y2. 

8. 

x6  +  y*. 

9. 

a* -14  a -f- 49. 

10. 

x2  -  (a  -  6)2. 

11. 

a2  —  (m  +  ?i)2. 

12. 

a2 -lis +  18. 

13. 

»2  +  4  a;  —  45. 

14. 

a2 +  13  a; +  36. 

15. 

a:2 -13  a: -48. 

16. 

a:2  +  9a; -36. 

17. 

x2-\-x-  110. 

18. 

2a;2  +  3z2/-22/2. 

19. 

s2-  6* -40. 

20. 

x4  +  x2y2  +  y4. 

21. 

a;2 -7a; -60. 

22. 

2a2  -la  +  6. 

•106  FACTORS. 

45.  a5  +  y5.  63.  ax*  +  £a8  -  as  -  J. 

46.  32  a5 -c5.  64.  (a-y)2-2y(a-y). 

47.  a6 +  64/.  65.  1  -  10  ay  +  25  a2y2. 

48.  729 -a6.  66.  a2  -  b2  +  2bc  -  c2. 

49.  a12-y12.  67.  a2  +  4  y2  -  z2  -  4  ay. 

50.  (a  +  6)4-l.  68.  a2  -  4  52  -  9  c2  +  12  be. 

51.  16a4 -81.  69.  4a2  +  9y2-z2-12ay. 

52.  a4  +  a2  +  l.  70.  (a  +  5)2  -  (c  -  d)2. 

53.  27a8 -64a8.  71.  a2  +  a  +  Sb-9b2. 

54.  a9  +  ?/9.  72.  5  c4 -15  c8 -90  c2. 

55.  a9  —  y9.  73.  a2a  —  c2a  +  a2y  —  c2y. 

56.  a9 -256.  74.  a4  +  16  a2x2  +  256  a\ 

57.  l-(a-y)8.  75.  (x  +  yy  +  (2x-yy. 

58.  a6 -216.  76.  a2  +  b2  -  c2  +  2ab. 

59.  a2 -4a -77.  77.  y2  -  a2  -  c2  -  2ac. 

60.  a*  +  bB  +  a  +  b.  78.  (a  +  5  a)2  -  25  a2. 

61.  a8-£8  +  a-6.  79.  2  ay  -  a2  -  y2  +  z2. 

62.  a2-62  +  a-£.  80.  4a4  -  9a2  +  6a  -  1. 

81.  a2-2a&  +  £2  +  12ay-4a2-9y2 

82.  2a2-4ay  +  2y2  +  2aa-2ay. 

83.  (a  +  b)2  -  1  -  ab  (a  +  b  +  1). 

84.  a8-a2  +  3a  +  5. 

85.  a2  —  y2  —  z2  —  2  yz  +  a  +  y  +  «. 

86.  a2  +  y2  +  *2-2ay-2a2  +  2y». 

87.  4a2#J-(a2  +  &2-c2)2. 


FACTORS.  107 

88.  x2  +  2x-3.  97.  x2  +  7xy  +  12y\ 

89.  a2  +  3a -40.  98.  »2  +  «2/-2y2. 

90.  a2-9z-10.  99.  x2  +  3a^-4?,2. 

91.  x4  +  8x2-9.  100.  a4-(?/  +  z)4. 

92.  a4-2z8-24z2.  101.  x8  -  y8  -  3  xy  (x  -  y). 

93.  z4-14z2-  51.  102.  sc8-2a:2  +  2-x. 

94.  x*-10ax  +  16a2.  103.  a;8-  8  -  6x  +  3z2. 

95.  ic4  -  15  x V  +  9  2/4.  104.  3  a2?/2  +  9  «y8  -  12  y4. 

96.  l-9a>-10a:2.  105.  a4  -  a8£  +  a&8  -  64. 

106.  a2-4c2  +  a-2c. 

107.  4ta2  +  9b2-c2  +  12ab. 

108.  15x2-5ax  +  3bx  —  ab. 

109.  3  a*  +  15  a»b- 24  a2b2. 

110.  6a8-30a26  +  36a&2. 

111.  25a2-4x2  +  4x-10a. 

112.  £c4y  —  sr^8  —  sc8y2  +  jc?/4. 

113.  9a2-462  +  3a  +  2&. 

114.  x8-y8-  (x2  -y2)-(x-  y)\ 

115.  (a._y)«_l_2(a>-y-l). 

116.  a8-2a2c  +  a2-4a  +  8c-4. 

117.  a2-&2-c2  +  26c  +  a. +  &-c. 

118.  x8z2  -  8  y8z2  -  4  x8n2  +  32  y87i2 

119.  5ac  +  3bc  +  c  +  5ab  +  3b2  +  b. 

120.  2  a&  -  2  5c  -  ace  +  ex  +  2  £2  -  bx. 

121.  s4  -  2  afa2  -  a4  -  a2£2  -  b\ 


CHAPTER  Vm. 
COMMON  FACTORS  AND  MULTIPLES. 


Highest  Common  Factor. 

137.  Common  Factors.  A  common  factor  of  two  or  more 
integral  numbers  in  Arithmetic  is  an  integral  number  that 
divides  each  of  them  without  a  remainder. 

138.  A  common  factor  of  two  or  more  integral  and  rational 
expressions  in  Algebra  is  an  integral  and  rational  expression 
that  divides  each  of  them  without  a  remainder. 

Thus,  6  a  is  a  common  factor  of  20  a  and  25  a. 

139.  Two  numbers  in  Arithmetic  are  said  to  be  prime  to 
each  other  when  they  have  no  common  factor  except  1. 

140.  Two  expressions  in  Algebra  are  said  to  be  prime  to 
each  other  when  they  have  no  common  factor  except  1. 

141.  The  greatest  common  factor  of  two  or  more  integral 
numbers  in  Arithmetic  is  the  greatest  number  that  will 
divide  each  of  them  without  a  remainder. 

142.  The  highest  common  factor  of  two  or  more  integral 
mnd  rational  expressions  in  Algebra  is  an  integral  and 
rational  expression  of  highest  degree  that  will  divide  each 
of  them  without  a  remainder. 

Thus,  3  et?  is  the  highest  common  factor  of  3  «2,  6  a8,  and  12  a4  ,- 
&x*y2  is  the  highest  common  factor  of  l@»*y2  and  15x2y2. 


COMMON  FACTORS  AND  MULTIPLES.  109 

For  brevity,  we  use  H.  C.  F.  for  highest  common  factor. 

1.  Find  the  H.  C.  F.  of  42  a%2  and  30  a2b\ 

±2a*b2  =  2  X  3  X  7  X  aaa  X  bb; 

30a2b*  =  2x3x5XaaX  bbbb. 

.'.  the  H.C.F.  =  2x3XaaXbb  =  6a2b2. 

2.  Find  the  H.C.F.  of  x2  -  9y2  and  x2  +  6xy  +  9y\ 

x2-9y2  =  (x  +  3y)(x-3y); 

x2  +  6*2/  +  %2  =  (a  +  3y)  (x  +  3y). 
.-.the  H.C.F.  =x  +  3y. 

3.  Find  the  H.C.F.  of 

'     4a;2-4a;-80;  2a:2-18a;  +  40;  2  a;2  -  24  x  +  70. 

4a;2  -  4*  -  80  =  4  (a2  -  a;  -  20) 

=  4(a-5)(a;-h4); 
2x2  -  18a;  +  40  =  2  (a;2  -  9x  +  20) 

=  2  (a;  -  5)  (a;  -  4)  ; 
2  a;2  -  24  a;  +  70  =  2  (x2  -  12  a;  +  35) 
=  2  (a;  -  5)  (x  -  7). 
.-.  the  H.C.F.  =  2  (a;  -  5).     Therefore, 

143.   To  Find  the  H.C.F.  of  Two  or  More  Expressions, 

Resolve  each  expression  into  its  prime  factors. 

The  product  of  all  the  common  factors,  each  factor  being 
taken  the  least  number  of  times  it  occurs  in  any  of  the  given 
expressions,  is  the  highest  common  factor  required. 

Note.  The  highest  common  factor  in  Algebra  corresponds  to  the 
greatest  common  measure,  or  greatest  common  divisor,  in  Arithmetic. 
We  cannot  apply  the  terms  greatest  and  least  to  algebraic  expres- 
sions in  which  particular  values  have  not  been  given  to  the  letters 
contained  in  the  expressions.  Thus  a  is  greater  than  a2,  if  a  stands 
for  \  j  but  a  is  of  lower  degree  than  a2. 


110  COMMON  FACTORS  AND  MULTIPLES. 

Exercise  48. 
Find  the  H.C.F.  of: 

1.  120  a2  and  168  a8.  4.   36  a*x2  and  28  x *y. 

2.  36  xB  and  27  x\  5.    48  a2b*c  and  60  a*c\ 

3.  42  a2x*  and  60  a*x2.  6.    8  (a  +  &)2  and  6  (a  +  ft)8. 

7.  12a(a;-f  y)2and4ft(a;  +  *,)8. 

8.  (a?  -  l)2  (x  +  2)2  and  (a;  -  3)  (x  +  2)8. 

9.  24  a2ft8  (a  +  ft)  and  42  a8ft  (a  +  ft)2. 

10.  sc2(ic-3)2  andcc2-3ic.    12.   x2-  4a;  and  a;2-  6a;  +  8. 

11.  a;2 -16  and  a;2 +  4  a;.         13.   a;2  -  7x  + 12  and  x2  -  16. 

14.  9»2-4y2and27a;8-83/8. 

15.  x2-  7x  -8  and  x2  +  5x  +  4. 

16.  £c2  +  3jcy-10y2anda;2-2«y~35y2. 

17.  a;4  -2a;8  -24a;2  and  6a;5-  6a4-  180a;8. 

18.  a;8-3a;2y  and  a;8- 27 y8. 

19.  l  +  64a;8andl-4a;  +  16a;2. 

20.  a;4  -  81  and  a;4  +  Sx2  -  9. 

21.  a;2  +  2x  -  3  and  x2  +  7x  +  12. 

22.  a;2  -  6a;  +  5  and  x2  -f  Sx  -  40. 

23.  3  a4  +  15  a%  -  12  a%2  and  6  a8  -  30  a2b  +  36  aft2. 

24.  Qx2y  -  12xy2  +  6y*  and  3ary  -f  9a;y8  -  12 y\ 

25.  l-16c4andl  +  c2-12c4. 

26.  9aa;8-aa;;  .9a;2- 6a; +  1;    27 a;8 -1. 

27.  x2-  3a;  -54  j   a;2 -a; -42;   a;2 -2a; -48. 

28.  8a;8  +  27y8;    4a;2  +  12a;y  +  9y2;   4a;2  -  9y2. 

29.  a;8  —  x2y  —  xy2  +  y8  j   x2  —  y2\   x2  -\-  2xy  -\-  y\ 


COMMON  FACTORS  AND  MULTIPLES.  Ill 

Lowest  Common  Multiple. 

144.  Common  Multiples.  A  common  multiple  of  two  01 
more  integral  numbers  in  Arithmetic  is  a  number  that  is 
exactly  divisible  by  each  of  the  numbers. 

A  common  multiple  of  two  or  more  integral  and  rational 
expressions  in  Algebra  is  an  integral  and  rational  expression 
that  is  exactly  divisible  by  each  of  the  expressions. 

145.  The  least  common  multiple  of  two  or  more  integral 
numbers  in  Arithmetic  is  the  least  integral  number  that  is 
exactly  divisible  by  each  of  the  given  numbers. 

The  lowest  common  multiple  of  two  or  more  integral  and 
rational  expressions  in  Algebra  is  an  integral  and  rational 
expression  of  lowest  degree  that  is  exactly  divisible  by  each 
of  the  given  expressions. 

We  use  L.  C.  M.  for  lowest  common  multiple. 

1.  Find  the  L.  C.  M.  of  42  a*b2 ;   30  aW ;   66  ab\ 

42a8b2  =  2  X  3  X  7  X  a8  X  J2; 
30aW  =  2  X3X  5Xa2X64; 
66  ab*  =  2  X  3  X  11  X  a  X  b\ 

The  L.  C.  M.  must  evidently  contain  each  factor  the  greatest  num- 
ber of  times  that  it  occurs  in  any  expression. 

.'.the  L.C.M.  =  2x3x7x5xllXa8X&4 

=  2310  a%\ 

2.  Find  the  L.C.M.  of 

4z2  -  4z  -  80  and  2x2  -  18a;  +  40. 
4z2  -  4aj  -  80  =  4  (x2  -  x  -  20)  =  4  (x  -  5)  (x  +  4)  ; 
2x*  -  18a  +  40  =  2  (x*  -  9x  +  20)  =  2  (x  -  5)  (x  -  4). 
.-.  the  L. CM.  =  4  (x  -  5)  (x  +  4)  (x  -  4).     Hence, 


112  COMMON  FACTORS  AND  MULTIPLES. 

146.  To  Find  the  L.  C.  M.  of  Two  or  More  Expressions, 

Resolve  each  expression  into  its  prime  factors. 

The  product  of  all  the  different  factors,  each  factor  being 
taken  the  greatest  number  of  times  it  occurs  in  any  of  the 
given  expressions,  is  the  lowest  common  multiple  required. 

Exercise  49. 
Find  the  L.  CM.  of: 

1.  x%y2  and  x2yz.  7.  abz,  a2b2c2,  and  abc*. 

2.  5  abc2  and  a2b2c\  8.  x2y,  xy2z2,  and  x2y*z. 

3.  4  xzy  and  12  xys.  9.  x2  and  x2  +  x. 

4.  5  a*bz  and  10  a2bB.  10.  x2  -  1  and  x2  -f  x. 

5.  21  xy*  and  28  x*y\  11.  a2  +  ab  and  aft  +  62. 

6.  10  x2z*  and  15  x2z2.  12.  a2+2a  and  (a  +  2)2. 

13.  a2  +  4a +  4  and  a2  +  5a -f  6. 

14.  a2  +  a;-20anda2-sc-30. 

15.  y2  -f  y  -  42  and  */2  -  11  y  +  30. 

16.  s2  -  10*  +  24  and  22  +  z  -  20. 

17.  (a  +  b)2;    (a-b)2',    a2  -  b2. 

18.  (a  +  2c)2;    (a  -2c)2;    a2  -4c2. 

19.  4ay(a;  +  2/)2;    2 x2  (cc2  -  y2)  ;   a8  (a  +  y). 

20.  a2  +  7a  +  12;   x2  +  6x  +  8;   x2  +  5x  +  6. 

21.  1-2/2;    1-2/8;    1  +  y. 

22.  a2  +  2icy  +  2/2;    x2-y2;    x2-2xy-\-y2. 

23.  a8-27;   a2  +  5z;   <c2  +  2a;-15. 

24.  y»-l;    2/8  +  2/2  +  y  +  l;    y8  -  y2  +  y  -  1.  . 

25.  (a-f-y)2-«2;    (x  +  y  +  z)2-,   x  +  y  -  z. 

26.  a;2-(ft  +  J)a;  +  ai;    cc2  —  (a  +  c)  a  -f-  ac. 

27.  aj2  +  3ay  +  2?/2;   a2 +-5a^  +  4y2;   x2-6xy-7y*. 

28.  jc2-7^  +  123/2;   s2  -  6a#  +  8y2;   x2-5^  +  6y3. 


COMMON  FACTOBS  AND  MULTIPLES.  113 

147.  The  chief  difficulty  in  finding  the  H.C.F.  and  the 
L.C.M.  of  two  or  more  algebraic  expressions  consists  in 
resolving  the  expressions  into  factors. 

When  two  numbers  in  Arithmetic  cannot  readily  be  re- 
solved into  their  prime  factors,  we  divide  the  greater  num- 
ber by  the  smaller;  then  the  divisor  by  the  remainder; 
and  so  on  until  there  is  no  remainder.  The  last  divisor  is 
the  greatest  common  factor. 

Likewise  in  Algebra  when  two  given  expressions  cannot 
readily  be  resolved  into  their  factors,  we  arrange  the  two 
given  expressions  in  descending  powers  of  a  common 
letter,  and  divide  the  expression  which  is  of  higher  degree 
in  the  common  letter  by  the  other  expression.  After  the 
first  division,  we  take  the  remainder  for  a  new  divisor  and 
the  divisor  for  a  new  dividend,  and  so  proceed  until  there 
is  no  remainder.  The  last  divisor  is  the  highest  common 
factor. 

Note.  If  the  two  expressions  are  of  the  same  degree  in  the  com- 
mon letter,  either  expression  may  be  taken  for  the  divisor. 

Find  the  H.C.E.  of  2x2  +  x  -  3  and  4a8  +  Sx2  -  x  -  6. 

2x*  +  x  —  3)4x*  +  8x2—    x-6(2x  +  3 
4*8  +  2a;2  —  6x 

6x2  +  bx  —  6 
6z2  +  3cc  —  9 


2z  +  3)2x2+    z-3(s-l 
2x*  +  Sx 

—  2x  —  3 

—  2x  —  3 


.'.the  H.C.F.  =  2z+3. 

Note.     Each  division  is  continued  until  the  first  term  of  the  re- 
mainder is  of  lower  degree  than  the  first  term  of  the  divisor. 


114  COMMON  FACTORS  AND  MULTIPLES. 

148.  This  method  is  of  use  only  to  obtain  the  compound 
factor  of  the  H.  C.  F.  Monomial  factors  of  the  given  expres- 
sions must  first  be  separated  from  them,  and  the  H.  C.  F.  of 
these  monomial  factors  must  be  reserved  to  be  multiplied 
into  the  compound  factor  obtained.  Also,  at  any  stage  of 
the  operation  a  monomial  factor  of  either  expression  may 
be  removed  without  affecting  the  compound  factor  sought. 

1.   Find  the  H.C.F.  of 

12a4  +  30a8  -  72a2  and  32a8  +  84a2  -  176a. 

12 aj*  +  30 «s  -  72 z2  =  6x2  (2x2  +  5x  -  12). 
32x3  +  84 X2  -  I76x  =  4x  (8x2  +  21x  -  44). 
6  x2  and  4  x  have  2  x  common. 

2x2+  6x  -  12)  8x2  +  21 «  -  44  (4 
8x2  +  20x-48 

x+    4)2x2  +  5x-12(2x-3 
2x2  +  8x 

-3x-12 
-3x-12 


.'.  the  H.C.F.  m  2x  (a  +  4). 

2.   Find  the  H.  C.  F.  of  4  a2  -  8  a  -  5  and  12a2  -  4a  -65. 

4x2-8x-5)12x2-    4x-65(3 
12x2-24x-15 
20x-50 

The  first  division  ends  here,  for  20  x  is  of  lower  degree  than  4  x2. 

We  take  out  the  simple  factor  10  from  20  x  —  50,  for  10  is  not  a 
factor  of  the  given  expressions,  and  its  rejection  can  in  no  way  affect 
the  compound  factor  sought,  and  proceed  with  2  x  —  6  for  a  divisor. 

2x-5)4x2-    8x-5(2x+l 
4x2  —  lOx 

2x  — 5 
2x-5 

.\theH.C.F.  =2a-5. 


COMMON  FACTOBS  AND  MULTIPLES.  115 

3.   Find  the  H.C.F.  of 

21a8  -  4a;2  -  15a  -  2  and  21a8  -  32 a2  -  54a  -  7. 

21x8-4x2-15x-2)21x3-32x2-54x-7(l 
21  x3—   4x2  — 15x  — 2 


28x2-39x-5 


The  difficulty  here  cannot  be  obviated  by  removing  a  simple  factor 
from  tlie  remainder,  for  —  28  x2  —  39  x  —  5  has  no  simple  factor.  In 
this  case,  to  avoid  the  inconvenience  of  fractions  we  multiply  the 
expression  21  x3  —  4x2  —  15x  —  2  by  the  simple  factor  4  to  make  its 
first  term  exactly  divisible  by  —  28  x2. 

The  introduction  of  such  a  factor  can  in  no  way  affect  the  H.C.F. 
sought,  for  4  is  not  a  factor  of  either  of  the  given  expressions ;  and  if 
we  multiply  only  one  of  the  expressions  by  4  we  do  not  introduce  a 
common  factor. 

The  signs  of  all  the  terms  of  the  remainder  may  be  changed  ;  for 
if  an  expression  A  is  divisible  by  —  F,  it  is  divisible  by  +  F. 

The  process  then  is  continued  by  changing  the  signs  of  all  the  terms 
of  the  remainder  and  multiplying  the  divisor  by  4. 

28x2  +  39x  +  5)84x3—    16x2—   60x—   8(3x 
84x3  +  117x2+    16x 

-133x2-   75x-   8 
Multiply  by  —  4,  —  4 

532x2  4-300x4-  32(19 
532  x2  4- 741x4- 95 
Divide  by  -  63,  —  63 )  -  441  x  -  63 


7x4-   1 

7x  4-  1)28x2  4-39x4-5(4x4-  5 
28x2+    4x 

35x  +  5 
35x  +  5 


.\theH.C.F.  =  7a  4-1. 


116 


COMMON  FACTORS  AND  MULTIPLES. 


4.   Find  the  H.C.F.  of 

8x2  +  2x  -  3  and  6x*  +  5x2  -  2. 


Multiply  by  4, 
8x2  +  2x  — 

Multiply  by  4, 

6x3  + 
4 

3)24x3  + 
24x3  + 

5x2-   2 

20x2-    8    (3x  +  7 
6x2—   9x 

14x2+    9x_   8 
4 

56x2  +  36x-32 
56x2  +  14x-21 

Divide  by  11, 

ll)22x-ll 

2a-    l)8x2  +  2x-3 
8x2  — 4x 

6x-3 
6x-3 

.the  H.C.F.  =  2  x 


The  following  arrangement  of  the  work  will  be  found 
jnost  convenient : 


3x 


8x2  +  2x-3 

6x3+    5x2—   2 

8x2  — 4x 

4 

6x-3 

24x3  + 20x2—    8 

6x-3 

24x8+    6x2—    9x 

14x2+    QX—    g 

4 

56x2  +  36x-32 

56x2 +  14x—  21 

ll)22x  —  11 

2x—    1. 

+  7 


4x  +  3 


Note.  From  the  nature  of  division,  the  successive  remainders  are 
expressions  of  lower  and  lower  degree.  Hence,  unless  at  some  step 
the  division  leaves  no  remainder,  we  shall  at  last  have  a  remainder 
that  does  not  contain  the  common  letter.  In  this  case  the  given 
expressions  have  no  H.C.F.  that  contains  the  common  letter. 


COMMON  FACTORS  AND  MULTIPLES.  117 

149.  In  the  examples  worked  out  we  have  assumed  that 
the  divisor  which  is  contained  in  the  corresponding  divi- 
dend without  a  remainder  is  the  H.C.F.  required. 

The  proof  may  be  given  as  follows : 

Let  A  and  B  stand  for  two  expressions  which  have  no 
monomial  factors,  and  which  are  arranged  according  to  the 
descending  powers  of  a  common  letter,  the  degree  of  B 
being  not  higher  than  that  of  A  in  the  common  letter. 

Let  A  be  divided  by  B,  and  let  Q  stand  for  the  quotient, 
and  B  for  the  remainder.  Then,  since  the  dividend  is 
equal  to  the  product  of  the  divisor  and  quotient  plus  the 
remainder,  we  have 

A  =  BQ  +  B.  (1) 

Since  the  remainder  is  equal  to  the  dividend  minus  the 
product  of  the  divisor  and  quotient,  we  have 

B  =  A-BQ.  (2) 

Now,  a  factor  of  each  of  the  terms  of  an  expression  is  a 
factor  of  the  expression.  Hence,  any  common  factor  of  B 
and  B  is  a  factor  of  BQ  -f-  B,  and  by  (1)  a  factor  of  A. 
That  is,  a  common  factor  of  B  and  B  is  also  a  common 
factor  of  A  and  B. 

Also,  any  common  factor  of  A  and  B  is  a  factor  of  A  — BQ, 
and  by  (2)  a  factor  of  B.  That  is,  a  common  factor  of  A 
and  B  is  also  a  common  factor  of  B  and  B. 

Therefore,  the  common  factors  of  A  and  B  are  the  same 
as  the  common  factors  of  B  and  B ;  and  consequently  the 
H.C.F.  of  A  and  B  is  the  same  as  the  H.C.F.  of  B  and  B. 

The  proof  for  each  succeeding  step  in  the  process  is 
precisely  the  same ;  so  that  the  H.  C.  F.  of  any  divisor  and 
the  corresponding  dividend  is  the  H.  C.  F.  required. 

If  at  any  step  there  is  no  remainder,  the  divisor  is  a  fac- 
tor of  the  corresponding  dividend,  and  is  therefore  the 
H.C.F.  of  itself  and  the  corresponding  dividend.  Heaee, 
this  divisor  is  the  H.  C.  F.  required. 


118 


COMMON  FACTORS  AND  MULTIPLES. 


150.  The  methods  of  resolving  expressions  into  factors, 
given  in  the  last  chapter,  often  enable  us  to  shorten  the 
work  of  finding  the  H.  C.  F.  required. 


1.   Find  the  H.C.F.  of 

a4  +  3a8 +  12a; -16; 


a;8 -13  a; +  12. 


Both  of  these  expressions  vanish  when  1  is  put  for  x.     Therefore, 
both  are  divisible  by  x  —  1,  §  135. 

The  first  quotient  is  x8  +  4  x2  +  4  x  +  16  =  (x2  +  4)  (x  +  4). 
The  second  quotient  is  x2  +  x  —  12  =  (x  —  3)  (x  +  4). 

Therefore,  the  H.C.F.  is  (x  -  1)  (x  +  4). 

2.   Find  the  H.C.F.  of 

2a;4  +  9a;8  +  14a; +-3;   3a;4  +  14a;8  +  9x  +  2. 


2x4  +  9xs  +  14x  +  3 


3x4  +  14x8  +   9x  +  2 
2 


6x*  +  28x8  +  18x  +  4 
6x4  +  27x3  +  42x  +  9 


x8  — 24x  — 5 
The  remainder,  x8  —  24  x  —  5,  vanishes  when  5  is  put  for  x. 
The  quotient  of  x8  —  24  x  —  5  divided  by  x  —  5  is  x2  +  5  x  +  1. 
Since  5  is  not  an  exact  divisor  of  3,  x  —  5  is  not  a  factor  of  2  x* 

+  9  x8  +  14  x  +  3 ;  but  x2  +  5  x  +  1  is  found  by  trial  to 'be  a  factor, 

and  is,  therefore,  the  H.  C.  F.  required. 

3.  Find  the  H.C.F.  of 

28a;2  +  39a +  5;   84 x*  -  16  jc2  -  60 x  -  8. 
By  §  132,  the  factors  of  28 x2  +  39x  +  5  are  7x  +  1  and  4x  +  5. 
The  factor  7  x  +  1  is  the  H.  C.  F.  required. 

4.  Find  the  H.C.F.  of 

2z4-6a8-a2  +  15a;-10;   4z4  +  6x*-  4z2-  15a;-  15. 


2x*-6x8-x2  +  16x-10 


4x*+   6x8-4x2-15x-15 
4x*  -  12 x8  -  2x2  +  30x  -20 


18x8-2x2 
The  remaindor  =  2  x2  (9  x  —  1)  —  6  (9  x  —  1) 

=  (2x2-5)(9x-l). 
The  factor  2  x2  —  5  is  the  H.  C.  F.  required. 


45x+   6 


COMMON  FACTORS  AND  MULTIPLES. 


119 


5.   Find  the  H.  C.  F.  and  the  L.  C.  M.  of : 

6  a;8  -  llx2y  +  2y*  and  9  a8  -  22  xy2  -  8y» 

6a8  —  llx*y  +  2y8 

6x8  —   8x2y  — 4xy2 

—  3x2y  +  4x£/2  +  2?/8 

-  3x2y  +  4sy2  +  2y8 


9x3  — 22xi/2  —    8y8 
2 

18  x8 -44x2/2- 16 1/8 
18x8-33x2y  +    6y8 

lly)33x2?/  —  44  xj/2  — 

22  y3 

3x2    -   4xy  — 

2y2 

.'.  the  H.  C.F.  =  3  a;2  -  4xy  -2y2. 


2x-y 


To  find  the  L.  C.  M.,  divide  each  of  the  expressions  by  the 
H.C.F. 

(6xs  -  llx2y  -{-  2 y*)  +  (3 x2  -  ±xy  -  2 y2)  =  2 x  -  y' 
(9  xs  -  22  xy2  -  8  f)  -s-  (3  x2  -  4  xy  -  2  y2)  =  3  x  +  4  y. 
. ' .  the  L.  C .  M.  =  (2  x  -  y)  (3  a;  +  4  y)  (3  a;2  -  4  ay  -  2  y 2). 


Exercise  50. 
Find  the  H.  C.  F.  and  the  L.  C.  M.  of : 

1.  4a;2  +  3a;-10;   4a;8  +  7a;2 -3a; -15. 

2.  2xs-6x2  +  5x-2;   8a;8  -  23  x2  -f  17  a;  -  6. 

3.  6a;8-7aa;2-20a2a;;   3  x2  +  aa;  -  4  a2. 

4.  3a;8  -  13a;2  +  23a; -21;    6 a;8  +  a;2  -  44 x  +  21, 

5.  c*-2c*  +  c,   2c4  -2c8  -2c  -2. 

6.  a8-6a2a;  +  12aa;2-8a;8;    2  a2  -  8  ax  +  8  a;2, 

7.  7a;8-2a;2-5;    7  a;8  +  12  a;2  +  10a;  +  5. 


8.    x*  -13  a;2  +  36;    a;4 


7a;2+.a;  +  6. 


9.  2a;8  +  3a;2-7a;-10;    4a;8  -  4a;2  -  9a;  +  5. 

10.  12a;8-  x2  -30a;  -16;    6  a;8  -  2a;2  -  13  a;  -  6. 

11.  6a;8  +  x2  -5x  -2;    6a;8  +  5a;2  -  3a;  -  2. 

12.  a;8-9a;2  +  26a;-24;    a;8  -  12  a;2  +  47  x  -  60. 


120  COMMON  FACTORS  AND  MULTIPLES. 

13.  4a;8 -2a;2 -16a; -91;   12 x8  -  28 x2  -  37 x  -  42. 

14.  a;4-4a:8  +  10a;2-12a;4-9;   a;4  +  2a;2-b9. 

15.  2  a;8 -3a;2 -16a; +  24;  4a;5  +  2x4-28x*-16x2-32x. 

16.  12a;8  +  4a;2 +  17 a; -3;    24a;8  -  52a;2  +  14a;  -  1. 

17.  2a;8  +  7aa;2  +  4a2a;-3a8;   4  a;8  +  9  aa;2  -  2  a2a;  -  a8. 

18.  2a;8-9aa;2  +  9a2a;-7a8;  4a;8- 20  aa;2  +  20  a2a;  - 16  a8 

19.  2a;4  +  9a;8  +  14a; +  3;    3a;4  +  14a;8  -f  9a;  +  2. 

20.  20a;8  +  2x2  -18  x  +  48  ;   20  a;4  -  17  a;2  +  48  a;  -  3. 

21.  2a;8  +  a;2-12a;  +  9;   2a;8  -  7a;2  +  12a;  -  9. 
2«.  a;8 -8a; +  3;   a;6  -  3a;5  +  21a;  -  8. 

23.  3x*-3x2y  +  xy*  —  ys;   4 a;8  -  a;2y  -  3 xy\ 

24.  8a;4  -  6a;8  -  x2  +  15a;  -25;   4a;8  +  7a;2  -  3a;  -  15. 

25.  4a;8 -4a;2 -5a; +  3;    10 a;2  -  19 x  +  6. 

26.  6a;4  -  13a;8  +  3a;2  +  2a;;   6a;4  -  10a;8  +  4a;2-  6a;  +  4 

27.  2a;4  -  3a;8  +  2a;2  -2x  -3;   4a;4  +  3a;2  +  4a;  -  3. 

28.  3a;4-a;8-2a;2  +  2a;-8;    6a;8  +  13  a;2  +  3  a;  +  20. 

29.  3a;6  +  2a;4  +  a;2;    3a;4  +  2a;8  -  3a;2  +  2a;  -  1. 

30.  3  -  2a;  +  5a;2  +  2a;8;    12  -  17a;  +  2a;2  +  3a;8. 

31.  10a; -6x2- 11a;8  +  9a;4- 6a;6; 

60a;  -f  4a;2  +  10 a;8  +  10 a;4  +  4 x\ 

32.  a;4-a;8-14a;2  +  a;  +  l;  a;5-  4  a;4  -  xz  -  2x*  +  8  x  +  2. 

33.  2a4 -2a8 -3  a2 -2a;   3a4  -  a8  -  2a2  -  16a. 

34.  6a;8-14aa;2  +  ^a2a;-4a8;  a;4-ax8-a2a;2-a8a;- 2a4. 

35.  4-2a;-8a;2  +  7a;8-9a;6;  2+ 5a;  -  10a;2- 7a;8  +  6a;4. 

36.  2a4  +  3a8a;-9a2a;2;    6 a4x  -  3 ax*  -  17 a8a;2  +  14 aV. 

37.  2a6  -  4a4 +  8a8  -  12a2+ 6a; 

3a«-3a6-6a4  +  9a8-3a*. 


COMMON  FACTORS  AND  MULTIPLES.  121 

151.  The  product  of  the  H.  C.F.  and  the  L.  CM.  of  two 

expressions  is  equal  to  the  product  of  the  given  expressions. 

Let  A  and  B  stand  for  any  two  expressions ;  and  let  F 
stand  for  their  H.C.F.  and  M  for  their  L.C.M. 

Let  a  and  b  be  the  quotients  when  A  and  B  respectively 
are  divided  by  F.     Then 

A  =  aF 
and  B  =  IF. 

Therefore,  AB  =  F  X  abF.  (1) 

Since  F  stands  for  the  H.  C.  F.  of  A  and  B,  F  contains 
all  the  common  factors  of  A  and  B.  Therefore,  a  and  b 
have  no  common  factor,  and  abF  is  the  L.  C.  M.  of  A  and  B. 

Put  M  for  its  equal,  abF,  in  equation  (1),  and  we  have 

AB  =  FM. 

152.  Since  FM=AB}  (§151) 

jtf==^  =  4x^  =  ^X  A.     That  is, 
F       F  F 

The  lowest  common  multiple  of  two  expressions  may  be 
found  by  dividing  their  product  by  their  highest  common 
factor,  or  by  dividing  either  of  them  by  their  highest  com- 
mon factor  and  multiplying  the  quotient  by  the  other. 

153.  The  H.  C.  F.  of  three  or  more  expressions  is  obtained 
by  finding  the  H.  C.  F.  of  two  of  them ;  then  the  H.  C.  F.  of 
this  result  and  of  the  third  expression  j   and  so  on. 

For,  if  A,  B,  and  C  stand  for  three  expressions, 
and  D  for  the  highest  common  factor  of  A  and  B, 
and  E  for  the  highest  common  factor  of  D  and  (7, 
then  D  contains  every  factor  common  to  A  and  Bf 
and  E  contains  every  factor  common  to  D  and  C} 
that  is,  every  factor  common  to  A,  B,  and  C. 


122  COMMON  FACTORS  AND  MULTIPLES. 

154.  The  L.  C.  M.  of  three  or  more  expressions  is  obtained 
by  finding  the  L.  C.  M.  of  two  of  them ;  then  the  L.  C.  M.  of 
this  result  and  of  the  third  expression ;  and  so  on. 

For,  if  A,  B,  and  C  stand  for  three  expressions, 

and  L  for  the  lowest  common  multiple  of  A  and  B, 
and  M  for  the  lowest  common  multiple  of  L  and  C, 
then  L  is  the  expression  of  lowest  degree  that  is 
exactly  divisible  by  A  and  B, 

and  M  is  the  expression  of  lowest  degree  that  is 
exactly  divisible  by  L  and  C.  That  is,  M  is  the  expression 
of  lowest  degree  that  is  exactly  divisible  by  A,  B,  and  C. 

Exercise  51. 
Find  the  H.  C.  F.  and  the  L.  C.  M.  of : 

1.  6x2  +  x-2',  2x2  +  7x-4:'J  2x2-7x  +  3. 

2.  a2  +  2ab  +  b2;  a2-b2;  a8  +  2a2b  +  2ab2  +  b\ 

3.  £c2-5aic+4a2;  x2-3ax  +  2a2;  3x*  -  Wax  +  1  a\ 

4.  x2  +  x-6-,  x*-2x2-x  +  2;  x*  +  3x*  -  6x  -  8. 

5.  a8-6a2  +  lla-6;   a8-8a2  +  19a;-12; 

x*-9x2  +  26x-24, 

6.  6x2  +  7xy-  3y2;    3x2  +  llxy  -  Ay2; 

2x2  +  llxy  +  12y2 

7.  8-14a  +  6a2;  4a  +  4a2-3a8;  4a2  +  2a8-6a4. 

8.  6cc8  +  7a;2-3a;  3x2  +  Ux-5;  6 x*  +  39 x  +  45. 

9.  27  a8 -a8;  6z2  +  aa;-a2;  15  a2  -  5  ax  +  3  bx  -  ab. 

10.  xz-l;2x2-x-l\3x2-x-2. 

11.  6a2-a-2;  21z2-17a  +  2;  14z2  +  5x-l. 

12.  12a2  +  2a;-4;  12x2  -  42*  -  24;  12  x2  -  28  x  -  24. 

13.  2aj2  +  3a;-5;  3cc2-a;-2;  2x2-\-x-3. 

14.  <e8  +  7a;2-l-5a;  -1;    a;2  +  3x  -  3a;8  -  1 ; 

3z8  +  5sc2  +  a;-l. 


CHAPTER  IX. 
FRACTIONS. 


Definitions. 

155.  An  algebraic  fraction  is  the  indicated  quotient  of 
two  expressions  written  in  the  form  -• 

156.  The  dividend  a  is  called  the  numerator;  the  divisor 
b  is  called  the  denominator ;  the  numerator  and  denominator 
are  called  the  terms  of  the  fraction. 

Fundamental  Principle  of  Fractions. 

Let  l  =  x.  (1) 

Multiply  by  b, 
Multiply  by  c, 

Divide  by  bcy  r~~x'  (?) 

From  equations  (1)  and  (2),  --  =  ~ 

o       be 

Now  —  is  obtained  by  multiplying  both  terms  of  -  by  c ; 
and  -  is  obtained  by  dividing  both  terms  of  -7-  by  e.    Hence, 

157.  If  the  numerator  and  denominator  of  a  fraction  are 
both  multiplied  by  the  same  number,  or  both  divided  by  the 
same  number,  the  value  of  the  fraction  is  not  altered. 


b 

=  X. 

a 

=  bx. 

ac 

=  hex. 

ac 
be 

=  X. 

124  FRACTIONS. 

Reduction  of  Fractions  to  their  Lowest  Terms. 

158.  A  fraction  is  in  its  lowest  terms  when  the  numera- 
tor and  denominator  have  no  common  factor.     Hence, 

To  Reduce  a  Fraction  to  Lowest  Terms, 

Resolve  the  numerator  and  denominator  into  their  prime 
factors,  and  cancel  all  the  common  factors  ;  or,  divide  the 
numerator  and  denominator  by  their  highest  common  factor. 

Eeduce  the  following  fractions  to  their  lowest  terms : 
38  a2b*c*      2xl9aW      2  6V 


57a8bc2       3x  19  a8bc2    '   Sa 


a 


(a  —  x)  (a2  +  ax  +  x2)  _  a2  -f-  ax  +  x2 


a2  —  x2  (a  —  x)  (a  +  x)  a  -\-  x 

a2  +  la  +  10       (a  +  5)  (a  +  2)       a  +  5 
a2  +  5a  +  6  ~  (a  +  3)  (a  -f  2)  ~  a  +  3* 
1 


•   a-3  _  2a;2  4.43.-  3 

We   find   by   the   Factor   Theorem  the  H.C.R   of  the 
numerator  and  denominator  to  be  x  —  1. 

The  numerator  divided  by  x  —  1      =  x2  —  3  x  -f- 1. 
The  denominator  divided  by  »  —  1  =  x2  —  x  +  3. 
x*-4x2  +  4x-l      z2-3a  +  l 


Therefore, 


:*-2x2  +  4x-3       x2-x  +  3 


Exercise  52. 
Reduce  to  lowest  terms  : 

6ab2  42m2&  34axV 

1#    9a2b'  4'    49 mn2'  '    51  a2xy7' 

m       3ab2c  30  xy*z*  Q     35aW 


3. 


15  aW  '    18  a2?/2*2  '     5a¥c 

26  a;  V  21  ^2^2  58  aW 

39  a:/'  6'    28 m2p'  '    87  aW* 


FRACTIONS.  125 

4a;2  +  12  ax  +  9  a2 


10.    —  -% zr~-'  19. 

12  xi  —  loxy 

Aa2  -9c2 

8a'  +  6a    ,  ^ 

a2  +  4a  +  4 

52-55 
13'   62-46-5*     ■  22' 

M'    4(a2  +  ac  +  c2)'  23' 

a;8  +  ?/8 
^5i    ^r 24. 

a;2  +  2  a;?/  +  y2 

16'    a;2  +  2a; -15*  25" 

17.    ^-^  +  11. 


■to      . .  07 

2a,2  -7a;  -15  4a2c2-(a2  +  c2-£2)2 


Eeduce  by  the  Factor  Theorem  j  or  by  finding  the  H.  C.  E 
of  the  numerator  and  denominator : 


a:2 

8  a;3  +  27  a8 
—  y2  —  2  yz  —  z2 

x> 

X* 

+  2  xy  +  y2  —  z2 
+  a;2?/2  +  y4 

x*  —  y 

2  a2  +  17 

3 

a +  21 

3  a2  +  26 
(a  +  by  - 

a +  35 
-c2 

(a 

(x 

+  6  + 

c)2 
-b2 

-6)2- 
+  a)2- 

-a2 
-b2 

(a 

+  6)2- 
+  b)2- 

-a2 

-(c  +  d)2 

(« 

+  c)2- 
0  + 

-  (b  +  d)2 
c)2  -  b2 

28-   3x*-8x  +  &'  32' 

x3  +  4ar  —  5 

3  xs  —  x2  —  x  —  1 
3a;3 -4a;2 -a; +  2* 

31.     ^-18«H-36  36. 


3a;8  +  17  a;2  +  22  a; +  8 


6a;8  +  25 

a;2 +  23  a; +  6 

a;8 

-3a;2 

-15. 

z  +  25 

X 

>  +  7a;2 

+  5a 

;-25 

2x*  +  x2 

-Sx 

+  3 

3a;8  +  8a; 

2  +  x 

-2 

a;8 

+  4a;2 

-Sx 

+  24 

a;4-a;8  +  8a;-8 


126  FRACTIONS. 

159.  There  are  three  signs  to  consider  in  a  fraction,  the 
sign  before  the  fraction,  the  sign  of  the  numerator,  and  the 
sign  of  the  denominator. 

s»"*>       i=—b  =  -— =-—b'         (§87) 

any  two  of  the  three  signs  may  be  changed  without  chang- 
ing the  value  of  the  fraction. 

The  sign  of  a  compound  expression  is  changed  by  chang- 
ing the  sign  of  every  term  of  the  expression.     Hence, 

1.  We  may  change  the  sign  of  every  term  of  the  numera- 
tor and  denominator  of  a  fraction  without  changing  the 
value  of  the  fraction. 

2.  We  may  change  the  sign  before  a  fraction  and  the  sign 
of  every  term  of  either  the  numerator  or  denominator  with- 
out changing  the  value  of  the  fraction. 

160.  From  the  Law  of  Signs,  therefore, 

1.  We  may  change  the  signs  of  an  even  number  of  factors 
of  the  numerator,  or  of  the  denominator,  or  of  both,  without 
changing  the  sign  of  the  fraction. 

2.  We  may  change  the  signs  of  an  odd  number  of  factors 
of  the  numerator,  or  of  the  denominator,  or  of  both  counted 
together,  if  we  change  the  sign  before  the  fraction. 

Eeduce  to  its  lowest  terms  -^ £4 — ; — Sr 

(b  —  a)  (c  +  d) 

Change  the  sign  of  the  factor  (b  —  a)  of  the  denominator  and  the 

sign  before  the  fraction,  and  we  have 

(q  —  b)  (c  —  d)  _       (a-b)(c  —  d)  _      c  —  d 

(b-a)(c  +  d)  {a-b){c  +  d)  c  +  d 

In  the  last  fraction  change  the  sign  of  the  numerator,  the  sign  of  the 

fraction,  and  the  order  of  the  terms  of  the  denominator,  and  we  have 

c  —  d _  d  —  c 

c  +  d~  d  +  c' 

Note.     Factors  and  terms  must  not  be  confounded. 


27  y>- 

-x3 

abx  — 

bx2 

ex2  — 

acx 

4ft2- 

6ab 

9  b2- 

4  a2 

1- 

X2 

4x  (x 

-1) 

<z2 +  5^-14 

8. 


9. 


FRACTIONS.  127 

Exercise  53. 

Keduce  to  lowest  terms : 

xy-Sy2  3ab(b2-a*) 

^(a^-ab2) 

c2-(a  +  b)2 
a2  -f-  ab  —  ac 

ac  —  be  —  ax  +  &b 

x'2  —  c2 

(b  +  c  +  d)2-a2 
(a-b)2-(c  +  d)2' 

(x  -  a)2  -  V2 
5*  4 -a2        '  '    (a-b)2-x2 

161.  Mixed  Expressions.     A  mixed  expression  is  an  inte 
gral  expression  and  a  fraction. 

Thus,  x  +  -  and  x  +  y tj—  are  mixed  expressions. 

x  x  "v  y 

Reduction  of  Fractions  to  Integral  or  Mixed 
Expressions. 

By  the  distributive  law  of  division,  §  43, 

a2  +  c      a2      c  ,   c 

= h  -  =  a-\ 

a  a       a  a 

Therefore,  a  fraction  whose  numerator  is  of  a  degree  equal 
to,  or  higher  than,  the  degree  of  the  denominator  is  reduced 
to  an  integral  or  mixed  expression  by  division.     Hence, 

162.  To  Reduce  a  Fraction  to  an  Integral  or  Mixed  Expression, 

Divide  the  numerator  by  the  denominator. 

mu      10s2  +  15x-7      10x2,   15  x       7        _      ,0       7 
Thus, =  — V —  =  2  x  +  3  —  - — 

'ox  ox         bx        ox  ox 


128 


FRACTIONS. 


Keduce   — — —  to  an  integral  or  mixed  expres- 


sion. 


a*  +  4  a2 
a8  +    a2 


2a 


a2  +  a  —  2 


a  +  3 


3  a2  +  2  a  —  5 

3  a2  +  3  a  —  6 

-    a  +  1 

The  remainder  —a  +  1  is  the  numerator  of  a  fraction,  and  the 
divisor  a2  +  a  —  2  is  its  denominator,  to  be  added  to  the  integral 
quotient  a  +  3.     Thus,the  complete  expression  required  is 


a  +  3  + 


a  +  1 


a2  +  a  —  2 


By  changing  the  sign  of  each  term  of  the  numerator  and  the  sign 
before  the  fraction, 

—  a  +  1  ,  «,  a  —  1 


a  +  3  + 


a  +  3 


a2  +  a-2  a2  +  a  — 2 

The  last  form  of  the  expression  is  the  form  usually  written. 


Exercise  54. 
Reduce  to  an  integral  or  a  mixed  expression : 


1. 


2. 


3. 


4. 


5. 


4x2-f  12x^3 


4a; 

3x2-6x- 

2 

3x 

x8  +  y8 

x  —  y 

xs  —  y8 

z  +  y 

as  +  2zs 

a-\-2x 


6. 


7. 


8. 


9. 


10. 


Ax2-3x  —  54 

x  —  4 

3  xs  —  x2  —  16  x  - 

-2 

x2  +  x-3 

X*  +  4:X*-7 

x2  +  x-2 
4  #2  +  6  ax  - 


21a' 


2x-3a 
5x8  +  9x*  +  5 


5  x2  •+  4  ic  —  1 


FBACTIONS.  129 

Reduction  of  Mixed  Expressions  to  Fractions. 

163.    The  value  of  a  number  is  unaltered  if  it  is  both 
multiplied  and  divided  by  the  same  number.     Hence, 
To  Reduce  a  Mixed  Expression  to  a  Fraction, 

Multiply  the  integral  expression  by  the  denominator,  to 
the  product  add  the  numerator,  and  under  the  result  write 
the  denominator. 

a2-ab-  b2 


Reduce  to  a  fraction  a  —  b  — 


a  +  b 


_  b  _  a1-^-  o2  =  (a  -b)(a  +  b)-  (a2  -ab-b2) 
a  +  b  a  +  b 


a 


2  +  ab  +  b2 


a  +  b 

ab 


a  +  b 


Note.  The  dividing  line  between  the  terms  of  a  fraction  has  the 
force  of  a  vinculum  affecting  the  numerator.  If,  therefore,  a  minus 
sign  precedes  the  dividing  line,  as  in  the  preceding  example,  and  this 
line  is  removed,  the  numerator  of  the  given  fraction  must  be  enclosed 
in  a  parenthesis  preceded  by  the  minus  sign,  or  the  sign  of  every  term 
of  the  numerator  must  be  changed. 

Exercise  55. 
Reduce  to  a  fraction : 

1.    a  +  b--^--  5.    a2  +  ax  +  x2 —  • 


a  +  b 

_  ,  a2  +  x2  n     a  —  Sx  ,   _ 

1,    a  +  x ■ 6. a  +  2  x. 

a  +  x  4 

3.   re  +  4 —  •  7.    3a  — 2b 


x  —  3  a  +  b 

xs 
a  +  x 


A       2             ,2          x*            o     o         9      21 -13  a? 
4.    a2  —  ax  +  xz ; 8.    2x  —  7 — 


130  FRACTIONS. 

Reduction    of    Fractions    to    Equivalent    Fractions 
Having  the  Lowest  Common  Denominator. 

1.   Reduce  - — 5 ;  tt~  »  tt~z  to  equivalent  fractions  having 
the  lowest  common  denominator. 

The  L.  C.  M.  of  the  denominators  is  12  a8.  (§  152) 

Divide  the  L.  C.  M.  by  the  denominators  4  a2;  3a;  6  a8. 
The  respective  quotients  are  3a;  4  a2;  and  2. 
Multiply  both  terms  of  the  given  fractions  taken  in  order  by  the 
respective  quotients,  3a;  4  a2;  and  2. 
We  have  for  the  required  fractions 

9ox     8a2y  10 

12  a8  '  12  a» '  and  12  a8 ' 

2-    ReduCG  a*  +  L  +  6'>  s'  +  4s  +  3  t0  e^ivalent  frac" 
tions  having  the  lowest  common  denominator. 

Express  the  denominators  in  their  prime  factors. 

2  3  2  3 


x2  +  5x  +  6  '  a2  +  4x  +  3      (x  +  3)  (as  +  2) '  (x  +  3)  (x  +  1) 
The  lowest  common  denominator  (L.  C.  D.)  is 

(x  +  3)  (x  +  2)  (x  +  I). 
The  respective  quotients  are  x  +  1 ;  x  +  2. 
The  respective  products  are  2  (x  +  1)  ;  3  (x  +  2). 
The  required  fractions  are 

2  (x  +  1)  3  (x  +  2) 

(x  +  3)  (x  +  2)  (x  +  1) ;  (x '+  3)  (x  +  2)  (x  +  1)' 

164.    Therefore,  we  have  the  following  rnle : 

Find  the  lowest  common  multiple  of  the  denominators  of 
the  given  fractions  for  the  common  denominator.  Divide 
this  common  denominator  by  each  of  the  given  denominators  ; 
and  multiply  the  given  numerators  each  by  the  correspond- 
ing quotient  for  the  required  numerators. 

Note.  Every  fraction  should  be  in  its  lowest  terms  before  the 
common  denominator  is  found. 


FRACTIONS.  131 


Exercise  56. 
Express  with  lowest  common  denominator : 

3a;-7     4o3-9  4  a2  +  c2     2a +  e 

r        6       ;       18  4d2-c2'    2a-c" 

a-2aj3a;2-2a  a2 -f  ?/2  1 

3a     '        9ax  25o32-4y2'  5o3  +  2y 

2o3-4y    3o;-8y  a;  +  2     a;  -  2 

3'       5cc2     '       10»  03-2*   03  + 2* 

4  a  — 5  c     3a  —  2c  4y2  xy 

4*       5ac      '      12a2c  *  3  (a  +  ft) '    6(a2-b2)' 

5            6  8x  +  2     2»-l      3o3  +  2 

5.  - :    z t,'  15. 


l-x>  l-x2  x-2'   3o3-6?    5o3-10 

1  2  a  —  bm      .      c  —  bn 

6*  o7+25  oT+T  16'       4a.     ;    1;      303 

i        a  a*  1  1 

7.  ;  -= ;•  17. 


.3 


(a-b)(b-c)>   (a-b)(a-c) 


1  X  X2  X8 

18. 


l+2a'   l-4a2  *  x-V   x  +  V   x2-l 

9           4  —  x  a  b  c 

9.  — s:  — 19. 


16  -a2'   4  +  03  *  a-b'   a  +  bJ   a2  -  b2 

a2             a  1  x  x 

0.  — -: 20 


27  -a*'  3- a  x  +  V    (z  +  l)2'    (x  +  l)2 

111 


21. 
22. 
23. 


(x-y)(x-z)'    (x-y)(x-c))    (x  -  c)  (x  -  z) 

1  1  1 

x2-5x  +  6'  ic2-4o3  +  3;  032-3oj  +  2* 

1  1 1 

o;  —  2a'   x2  —  5  ax  -\-  6  a2i  x  —  3  a 


132  FRACTIONS. 


Addition  and  Subtraction  of  Fractions. 

L    Find  the  algebraic  sum  of  -  H 

a       b       c=a+b-c 

XXX  X 

2.   Find  the  algebraic  sum  of 

3a  —  4#       2a  —  b  -\-  c      a  —  kc 
4  3  +  ~~ 12 

The  L.  C.  D.  =  12. 
The  multipliers,  that  is,  the  quotients  obtained  by  dividing  12  by 
4,  3,  and  12,  are  3,  4,  and  1,  respectively. 
Hence,  the  sum  of  the  fractions  is 

9a-126      8a-46  +  4c      a-4c 
12  12  12 

_0a  — 12  6  —  8a +  46  —  4c  +  a  —  4c 

12 
_  2a  — 86  — 8c_  a  — 46  — 4c 
12  ~  6 

The  preceding  work  may  be  arranged  as  follows  : 

The  L.  C.  D.  =  12. 

The  multipliers  are  3,  4,  and  1,  respectively. 

3  (3 a  —  4 b)      =      9a  — 12b  =  1st  numerator. 

—  4  (2 a  —  b  +  c)=  —  8a  +    4 6  —  4 c  =  2d  numerator. 

1  (a  —  4 c)         =       _a —  4c  =  3d  numerator. 

2a—    86  — 8c 
or  2  (a  —  4  b  —  4  c)  =  the  sum  of  the  numerators. 

.,      ^  2(a  — 46  — 4c)      a  — 46  — 4c 

.*.  sum  of  fractions  =  — i — '-  = • 

12  o 

165.  To  Add  Fractions,  therefore, 

Reduce  the  fractions,  if  they  have  different  denominators, 
to  equivalent  fractions  having  the  lowest  common  denomina- 
tor ;  and  write  the  algebraic  sum  of  the  numerators  of  these 
fractions  over  the  common  denominator. 

Note.  The  resulting  fraction  should  be  expressed  in  its  lowest 
terms. 


FRACTIONS. 


133 


Exercise  57. 
Find  the  algebraic  sum  of : 

x  —  1      x  —  3      a;  —  7      a;  —  2 
2  5  10    +     5 

2x  —  1       0  +  7      0  —  4      a;  —  3 
2"         3       ~      6  4      ~~      2     ' 

70-5  _  30  +  2      0  +  1  _  5  a;  -10 

8  3  4  12 

20  +  3   ,0  —  2      50  +  4      20-4 

9  +      6  12  3 

20  +  3   ,  0  +  3       180  +  5      x  -  3 


3. 


4. 


5. 


2x 


4a; 


80s 


x 


x      2x  -11       a; +  3      a;  — 7      a;  -  1 
6*    2  3  4     +     6  12    ' 


4a_2_  q  +  5       4&»      a26  +  ab2  -  4  a4 
£2  «6  aa  +  a262 

5  a; -11  _  0-1       11  a; -1  _  12  a; -5. 
8'  4  10  12  3 

0  +  1    ,   30-4   ,   1       60  +  7 


9. 


10. 


11. 


2       '        5       +4  8 

2-0-6      80-4  .  56a; -48 


50 


15  0 


45  0 


11 0y  +  2      5y2-3      602-5 


x*y2 


xif 


xsy 


a  —  b   ,  b  —  e      c  —  a   ,   a£>2  +  &c2  +  ca2 
cab  abc 

1         2x  —  g      y— 60 
2az2y  '   6y2«      2  0s2        4aV        ±x2yz 


13.    JU     1 


134  FBACTIONS. 

166.  When  the  denominators  are  polynomials  arranged 
in  the  same  order,  we  first  express  the  denominators  in 
their  prime  factors. 

Find  the  algebraic  sum  of : 

a  4-  b      a  —  b         Aab 
a  —  b      a  4  b      a2  —  b2 
as  -  &2  =  (a  +  6)  (a  _  &). 
The  L.C.D.  is  (a  -  6)  (a  +  6). 

The  multipliers  are  a  4  6,  a  —  6,  and  1,  respectively. 
(a  4  6)  (a  4  b)  =      a2  4  2  a&  +  62  =  1st  numerator. 

—  (a  —  b)  (a  —  b)  =  —  a2  4  2  ab  —  b2  =  2d  numerator. 

—  1  (4  a&)  =  —4ab  =  3d  numerator. 

0  =  sum  of  numerators. 


.*.  sum  of  fractions  =  0. 

Exercise  58. 
Find  the  algebraic  sum  of : 

1-f  x 


4 z-  8. 


x+6      x-5  1+x  +  x2      1-x  +  x2 

2  __1 L_.  9  a  +  y     g-y       ±xv  . 

'  l+a;       1  —  a?  '  x  —  y      x -\- y      x2  —  y2 

3,l+a;       1  -  *•'  10'  a  -  a  +  a  +  SB  +  a2  4  z2* 

c  a  +  y       s*  -  y2  n    1          2       |       1     , 

'  x  —  y      (x  —  y)2  "  x      x  4  1      x  +  2 

x-y      (x-y)2  6-2a  2 1_ 

a?  +1/       (a  +  y)2  ■    9 -a2       34  a      3- a 

6       i.i.    13  _^__^ «*!_ 

2a24-2a^"1r2a2-2a6  a4&      (a  +  i)2      («46)8 

1            Q-3c)2  s  +  y        2a;        x2(x-y) 

a-3c      a*-21c*  '      y        x+y      yi^-y2) 

Hint.     Reduce  the  last  fraction  to  lowest  terms. 


FRACTIONS. 


135 


15. 


x  +  2y      x  —  2y      x2  —  4  ya 
1  1 


16.    -L-?  + 


17. 


18. 


19. 


y  —  1       y      y  -{  y2  - 

x 


a;'5  —  x      x 
4  a  5  a2 


x  -1 

3 

x  —  a       (x  —  a)2       (x  —  a)8 

1  a2  x  +  a 


x-2a      xs-8a*      x2  +  2ax  +  4:a* 

1 


.„     «2-2a;  +  3,        x-2 
20.   s-^— : f- 


OJ8-hl 


ar  —  x+1      x  -f- 1 


21. 


22. 


1       .         £C-1         ,  a2  +  a  -  3 

+      0.0  .     r,  +  — 1 


»-3ic2 +  3^4-9         x8  -  27 
x2  +  8  x  +  15      x  -  1 


x2  +  7x-f-10      a; -2 
Hint.     Reduce  the  first  fraction  to  lowest  terma 

x2  —  5  ax  +  6  a2       x  —  1  a 


23. 


24. 


ic2  —  8  ace  +  15  a2      x  —  5  a 
11  2 


a  -2      x2-3x  +  2      x2-4x  +  3 


Hint.     Express  the  denominators  of  the  last  two  fractions  in 
ime  factors. 


25. 
26. 
27. 


a2-7a  +  12      a2-4a-f-3      a2-5a  +  4 

3 4 

10a2  +  a -3      2a2  +  7a -4* 

3 1 

2-x-6x2      \-x-2x2 


136  FRACTIONS. 

167.  When  the  denominators  are  polynomials  not  ar- 
ranged in  the  same  order,  we  first  write  the  fractions,  by 
§  159,  so  that  the  denominators  shall  be  arranged  in  the 
same  order. 

1     tk  a  <u  *2  3        L  2*-3 

1.  Find  the  sum  of - 7  +  z : — ;• 

x      2x  —  1       1  —  4x2 

Change  the  signs  before  the  terms  of  the  denominator  of  the  third 
fraction  and  the  sign  before  the  fraction.     We  now  have 
2_       3  2x-3 

x      2x  — 1      4x2  —  1* 
Proceeding  as  in  §  166,  we  find  for  the  required  sum 

-2 2 

x(2x— l)(2x  +  l)      x(l-2x)(l+2x)' 
if  we  change  the  sign  of  2  and  the  signs  of  (2  x  —  1),  §  160. 

2.  Find  the  sum  of 

1      i      * 


a  (a  —  b)  (a  —  c)      b(b  —  a)  (b  —  c)      e(c  —  a)  (c  —  b) 

Change  the  sign  of  the  factor  (b  —  a)  in  the  denominator  of  the 
second  fraction,  and  change  the  sign  before  the  fraction,  §  160. 

Change  the  signs  of  the  two  factors  (c  —  a)  and  (c  —  b)  in  the 
denominator  of  the  third  fraction,  §  160.     We  now  have 

* ! + 1 

a(a  —  6)  (a  —  c)      b{a  —  6)  (6  —  c)      c(a  —  c)(b  —  c) 
The  L.C.D.  =  abc  (a  -b)(a-  c)  (b  -  c). 

bc(p  —  c)  =  62c  —  6c2  =  1st  numerator. 

—  ac(a  —  c)  =  —  a?c  +  ac*  =  2d  numerator. 

afr(a  —  b)  =  a2b  —  ab2 =  3d  numerator. 

a2b  —  a?c  —  ab2  +  ac2  +  b2c  —  be2  =  sum  of  numerators. 
=  a2(b-c)-a (b2  -  c2)  +  bc{b-  c). 
Divide  by  the  common  factor  (6  —  c),  and  we  have 
a2  —  ab  —  ac  +  be  \ 
and  this  is  equal  to  (a  —  b)  (a  —  c). 

.-.  (a  —  6)  (a  —  c)  (6  —  c)  =  sum  of  numerators. 

tt     +.  (a-b)(a-c)(b-c)    _  1  m 

,.  sum  of  fraction*         "  «bc(a-6)<a-c)^-c)  "  o6c 


FRACTIONS.  137 


Exercise  59. 
Find  the  algebraic  sum  of : 


—  1      x  +  1      1  —  x 
a  Sa  2  ax 


a  —  x      a-{-  x      x2  —  a2 
3      _       2  !5 


2a-3      3  +  2a      9-4a2 

c&  —  b         2a         a3  +  a26 
T~~*"a~^b  +  b*-  a2b 

3  5  2z-7 


x      l-2x      Ax2-1 


(x  H-  a)2      (a  —  #)2      £p2  —  a 

7#        1  s-y  a:y-2a;a 

'   x  —  y      x2  -\-  xy  -\-  y2        y8  —  x* 

8.    ,  *         *  4 


(a;  -  2)  (*  -  3)      (aj  -  1)  (3  -  as)      (x  -  1)  (as  -  2) 

be ac ab 

(c  —  a)  (a  —  b)       (a  —  b)  (b  —  c)       (b  —  c)  (c  —  a) 

b  -f  c a  +  c  ,  a  +  b 


(a  —  b)  (a  —  c)       (b  —  c)  (b  —  a)       (c  —  a)  (c  —  b) 

3 4 6 

(a  —  b)(b  —  c)       (b  —  a)  (c  —  a)       (a  —  c)  (c  —  b) 

I + l i 

x(x  -y)(x-  z)      y(y  -  x)  (y  -  z)      xyz 

13  f£zj£  b2  +  ae  e2  +  ab 

(a  —  b)  (a  —  c)       (b  —  a)  (b  4-  e)       (e  —  a)  (c  +  b) 


11. 


12. 


138  FRACTIONS. 


Multiplication  of  Fractions. 
Find  the  product  of  -  X  -v 

Let  ixi  =  *-  W 

Multiply  each  of  these  equals  by  b  X  d. 

Since  the  order  of  the  factors  is  immaterial,  (§  42) 

(  f  X  M  X  (^Xdj=bXdXx. 
Or,  aXc  —  bXdXx. 

Divide  by  bxd,  f£j=«-  V) 

From  (1)  and  (2),  ^X°-  =  ;~ 

T ...       .  a      c       e       a  X  c      e       a  X  c  X  e 

Likewise,         -  X  -  X  -  =  - ;  X  - .—  ; ; - ; 

b       d      f      bXd      J       bXd  Xf9 

and  so  on  for  any  number  of  fractions.     Hence, 

168.   To  Find  the  Product  of  Two  or  More  Fractions, 

Find  the  product  of  the  numerators  for  the  required 
numerator,  and  of  the  denominators  for  the  required  de- 
nominator. 

In  applying  the  rule,  reduce  every  mixed  expression  to  a  fraction, 
and  every  integral  expression  to  a  fraction  with  1  for  the  denominator. 

Cancel  every  factor  common  to  a  numerator  and  a  denominator,  as 
the  cancelling  of  a  common  factor  before  the  multiplication  is  equiva- 
lent to  cancelling  it  after  the  multiplication. 

4     _.    ,  J  ,  .  2  a%      6  cH       5  ab*c 

1.   Find  the  product  of  —  X  —  X  g-j^ 

2a?b      QcH       bab^c  _  2  X  6  X  5aWc*d  _  ^  , 
3  cd*      5  aft      8  a  W      3X5X8  aWd*      2  d»  * 


FRACTIONS.  139 


2.   Find  the  product  of 

x2  -  y2  xy-2y2      x2  -  xy 

x2-3xy  +  2y2       x2  +  xy        (as  -  y)2 

Express  the  numerators  and  denominators  in  prime  factors. 
(x  -  y)  (x  +  y)  xy(x-2y)  x  x(x-y)  _  y  _ 
(x  —  y)(x  —  2y)       x(x-\-y)        (x  —  y)(x  —  y)      x  —  y 

The  common  factors  cancelled  are  (x  —  y),  (x  +  y),  (x  —  2  y),  x. 
and  (x  —  y). 


Exercise  60. 

Find  the  product  of : 

12  x2      Uxy 
1.    -=-?  X  r-rj-  10. 

7  y2       9  ars8 


3a%2c*      20  mV 
2.    -j X  —     .     •  11. 

4  aw»        21  arc6 

7  rascy        3  asc* 

9mV       4  a2?/2 

8  x8y*       W  mn 

21«y     4a2b 

6.  J*K  x  Mtf.  15. 

12  «2      35ic« 

„    x2  —  y2        4  x 

7.  2  ,     o  X  — 16. 

cc2  +  y2      x  +  y 

o     3cc2-a  2a 

8.  ■ — *7n — a —    17- 

a  2  x2  —  4  x 


3x*  3x-6 

5* -10*     4x8 


18. 


a8  -  j«  A 

±a2 

x2  4-  2/2  x 
a;2  —  y2 

x  +  y 
3x2+  3y2 

ab-b2 

a  (a  -\r  I) 

a(a2-b2) 
b2 

a2- Ax2 
a2  4-  4  ax 

ax  4-  4  x2 
a2  —  2  ax 

x*  ~~  yA      x  ~  y 

(x  —  y)2  *'\  x2  4-  xy 

x*  +  a* 
x2-9a2 

w  a?  4-  3  a 
jc  4-  a 

a*-b* 
a*  -f-  b*  X 

a2-ab+  b2 
a  —  b 

x2-l 

x>  -25 

x2  —  4:X  — 

5"x8+2a;-3 

(,-£\ 

x  f   ."*    Y 

140  FRACTIONS. 


fa.  _  xy  ~  y2\  (x  -  xy2  ~  y*\  x      *2 

\         x  +  V  )  \         x2  +  y* )      x2  —  xy  +  y2 


21 


.     8  a2b       c2d       4  ab       bed  —  cd2 
22.    X  7T-z  X  — -  X 


8a8        co"        ±(b2-bd) 


23   y(**-y*)  x    fr2-*/2)2    x  fr  +  y)2 

x(x  +  y)       x2  +  xy  +  y2      (x  —  y)2 

■        (a  +  b)2-e2  x  a  x  (a-b)2-e2 

a2  +  aft"  —  ac       (a  -f  c)2  —  ft3      ab  —  b2  —  be 


G 


(a  +  b)2-e2      c2-(a-b)2 

a2-(b-  e)2      c2-(a  +  b)2      ae  -  a2  +  ab 

(x  -  a)2  -  b2      x2-(b-  a)2      ax  +  a2-  ab 
(x  -  bf  -  a2  X  x2  -  (a  -  b)2  X  bx-ab  +  b2 


27. 


a2-2ab  +  b2-c2      a  +  b 


a2  +  2ab  +  b2-c2      a-b  +  e 


x*  +  (x  +  1y  X2  +  X  2x  +  l 

X  (x  +  1)  (x  +  l)2  -  x2  x 

2  axs  4-  2  a8x  x2  —  a2        x  +  a 


29. 


(x  -  a)2  (x  4-  a)2      2  (x2  4-  a2)         ax 

^    a2-b2-e2-2be      (.  2c       \ 

a£  —  ab  —  ac  \        a  +  b  +  ej 

aA  +  a2b2  +  V        a  +  b       a2  -  b2 
!r  *•-*•        Xa84-ft8><       a 

a;24  7o;y  +  12y2       a:3  +  ay  -  2 y8 


FRACTIONS.  141 


Division  of  Fractions. 

169.  Reciprocals.     If  the  product  of  two  rmmbers  is  equal 
to  1,  each  of  the  numbers  is  called  the  reciprocal  of  the  other. 

The  reciprocal  of  -  is  - ;   for  -  X  -  =  -7  =  1. 
0      a  0      a      ao 

The  reciprocal  of  a  fraction,  therefore,  is  the  fraction 
inverted. 

Find  the  quotient  of  -  ■+■  -• 

Let  §  +  |  =  *  (1) 

Since  the  dividend  is  the  product  of  the  divisor  and 
quotient, 

a      c 

Multiply  each  of  these  equals  by  -• 

mi  a      d       c      d  _„ 

Then  -  X  -  =  -  X  -  X  x  =  x.  (2) 

0      c      a      c  v  ' 

From  (1)  and  (2),  \  -*-  4=  T  *  -'     Therefore, 

w  v  *    b      d       b       c 

170.  To  Divide  by  a  Fraction, 

Multiply  the  dividend  by  the  reciprocal  of  the  divisor. 

Fmd  the  result  of  -  X  -2 : z  h « — 7^ — 

x      x*  —  4  a;  —  5  xr  —  25 

1  x2-l        .  gg  +  2a?-3  =  l  x2-l  x2-25 

x     x2  — 4x  — 5  *       x2  —  25  a;      x2  — 4x  — 5      x2  +  2x  — 3 

,1      (x-l)(x  +  l)      (g-5)(g  +  5>==    x  +  5 
x      (x-5)(x  +  l)      (x  +  3)(x-l)      x(x  +  3) 
The  common  factors  cancelled  are  (x  —  1),  (x  +  1),  (x  —  6). 


i 


142  FRACTIONS. 

Exercise  61. 

Find  the  quotient  of: 

15a2  .  9x»z  5c^      3b»       She 

'     If    :  2Sxy'  2abX  5ac  ''  2a9' 


2. 


3x2y2z2  m   18  x2yz2  x2-a2   %.    x  -  a 

±a2b2c2  :    9a*b2c'  x2  -la2'*'  x  +  2a 

5oW    .  20 'nMp*  x2y2  +  3xy  ,  xy  +  3 

3m*np*  '     2±abc*  '  4c2-l      :  2c-l 

16  a  W        8  aW  9x2~4y2      3x-2y 

21  m2xY  :  7m*xy'  9*       4  -  x2     ~*~     2+x 

2& ae  y  a&  a2  —  4       a2  +  2  a 

fo    !  3£2      5c2'  *      a2  +  4a~%2-16* 

ft2-4a  +  3      a2-10a  +  21      a2-7a 


11. 


12. 


a2-5a  +  4,'    a2  -  9  a  +  20       a2  -  5  a 

b2-7b  +  6  .  £2 -145 +  48      &2  +  66 
62-f  36-4  :  62  +  10&  +  24  :  b*-8b2' 


13  a;2-y2  %2+-a;y        x2  -  xy 

x2-3xy  +  2y2  '  xy-2y2      (x-y)* 

(a -by      2ab-2b2      a?  +  ab 
a2-b2    '  3  X~a~=T" 


15. 
16. 
17. 
18. 


(a  +  by-c2  m  c2-(a  +  b)* 
a?-(b-  c)2  '  c2-(a-  b)2' 

(x-ay-b2  m  x2-(a-b)2 
(x-b)2-a2  *'  x2-(b-ay 

(a  +  by -(c  +  d)2  m  (a- cy -(d- by 

(a  +  cy-(b+d)2  '   (a-b)2-(d-cy 

x2  —  2  xy  +  y2  —  z2      x  —  y  -\-z 
x2  +  2xy  -f  y2  —  z2  '  x  +  y  —  z 


FRACTIONS.  143 


Complex  Fractions. 

171.   A  complex  fraction  is  a  fraction  that  has  one  or 
more  fractions  in  either  or  both  of  its  terms. 

3x 

Simplify  the  complex  fraction  •? — ^-r  • 

4 

3«  0      .  4  s  —  1       0     '         4  12 x         _ 

=  3  x —  =  3  x  X = •    Hence, 

4x  —  1  4  Ax  —  1      4x  —  1  ■ 


172.  To  Simplify  a  Complex  Fraction, 

Divide  the  numerator  by  the  denominator. 

173.  The  shortest  way  to  simplify  a  complex  fraction  is 

to  multiply  both  terms  of  the  fraction  by  the  L.  G.  D.  of  the 

fractions   contained   in   the  numerator   and   denominator, 

§157. 

a  a 

.,     a.      ,.„    a  —  x      a  +  x 
1.   Simplify 


a  +  x 


The  L. CM.  of  a  —  x  and  a  +  x  is  (a  —  x) (a  +  x). 
Multiply  both  terms  by  (a  —  x)  (a  +  x),  and  we  have 
a  (a  +  x)  —  a  (a  —  x)  _  a2  +  ax  —  a2  +  ax  __2ax  _ 
x(a-{-  x)  +  x(a  —  x)~  ax  +  x2  +  ax  —  x2~  2ax~ 

2.  Simplify =■  • 


1  + 


1  + 


Multiply  both  terms  of  the  last  complex  fraction  by  1  —  x.     We 

\ x 

e ?  and  this  put  in  place  of  the  last  complex  fraction  changes 

A        X  1 

given  fraction  to  the  form 


i+i. 


2 x 

Multiply  both  terms  of  this  fraction  by  2  —  x.     We  have 


3-2 


144  FRACTIONS. 

Exercise  62. 
Simplify : 


1. 

£  +  £ 
m      m 

z 

m 

2. 

z 

X 

3. 

^-Sd 
c 

ab 

m  +  n 
1 


c  — 


8. 

2m  -f-  n       . 
m  ■+■  n 

1           n 

9. 

m  •+-  n 

xs +  y* 
x2-y2 

x2  —  xy-\-  y2 

x-y 

1               a 

o 

a-b      a2-b2 

a                 b 

1+-^  4+A+1 


11. 


ab  +  b2      a2  +  ab 

_I  +  1  +  I 

ab      ac      bo 


+  c)s 


x  +  1  a# 

5.  _^i^.  i2. 2  »  c, 

i 2>_  -  +  -  +  - 


x  +  y  b      c      a 

mn 

m  -\ 

m  —  n 

6. 13. 

mn 
m 


14. 


o  x  —  1 


1 

1 

1 

ab 

ac 

be 

a2- 

-(«- 

•c)2 

a 

1 

x2- 

-1 

x+- 

1 

FRACTIONS. 


145 


15. 


16. 


x  +  y 


a  + 


1  + 


a  +  1 
3-a 


x  +  y  + 


x-y  + 


x+y 


x  + 


17. 


x  + 


r-M 


y  (xyz  +  x  4-  *) 


18. 


3<z6c 


#c  4-  ac  4-  «6 


ft-1      5-1      c-1 

a  b  c 

a       o      c 


20. 


21. 


(m  +  n      m2  4-  w2\    m  /m  —  w      m2  +  w2\ 
m  —  n      m2  —  t&2/    '   \ra  +  n      m2  —  n%J 

fx-y  _  &  -  y*\  x  fn  +  y  +  ^_±_f\ . 

\*  +  2/      x*  +  y*J       \x-y      x2  -  y2J 


1  1 


a      6  4-c 


22. 


14- «  4 


1  —  a;  4-  x' 


23. 


x2  —  y2  —  z2  —  2yz 

x2  —  y2  —  z2  +  2yz 

x-y-z 

x  +  y  —  z 


146  FRACTIONS. 

Exercise  63.  —  Keview. 
When  a  =  2,  b  =  —  2,  and  c  =  4 ;   find  the  value  of : 


1.    VV  +  b*  +  c8  -  (a  -  b  -  c)\ 

a2  +  b2-c2  +  2ab 
•    a2_b2_c2  +  2bc' 

3.  (b  -  5)  (J  -  4)  -  3  (b  -  2)  (b  -  1)  +  3  (b  +  1)  (6,+  2). 

4.  Eednce  to  lowest  terms  ^  +  13x2  +  17x  +  6' 
Simplify : 

l  x— 4  «  — a 

5-  "3 ^t*t «c  —  z tt-; ;t:  + 


(x-S)(x-2)       (x-l)(x-3)   '    (a;-l)(a;-2) 
a  a  2  a2  4  a4 


'    a-&       a +  6       a2  +  b2      a4  +  64 

Hint.     Add  the  first  two  fractions ;  then  their  sum  and  the  third 
fraction  and  this  result  to  the  fourth  fraction. 


7.    ~z ^--^-7-tH 


a1 -2      a2 +  2      dz  +  l 


8      1     •  {(   1     I     1  V     2    V 

x2-2x  +  \        (x2-l      x  +  2\ 
(x-2)2      ''    \x2-±X  x  +  1) 


10. 


11. 


a2-b2  ab-2b2  ^_  (a  -  b)2 

a2-3ab  +  2b2><    a2  +  ab    '  a{a-b) 

(a  +  b)2  -  c*  a2b2c2  W 

a2  +  ab  —  ac      a2  -+-  ab  +  ac      a#c 


s2-f7a:y  +  10y2  s  +  1         .       1 

'    x2  +  6xy  +  5y*       x2  +  4a  +  4  *  x  +  2 


FRACTIONS.  147 

13  x2-\-yz  ,  y2jtxz        .  z2  +  xy 

(x-y)(x-z)       (j/-z)(y-x)       (z-x)(z-y) 

14.   (    x      ^"^W    ^         i  — A. 
\l  +  a;  x     J   '    \l-\-x  X     J 

c\x  —  c      x  +  2c)      x2  +  ex  —  2  c2 

16.  *-*(j* 1 1    £—) 

x2y2    \x2  —  y2  x2  -\-  y2J 

3(x2  +  x-2)  _  3(x2-x-2)  _      Sx 
x2-x-2  x2  +  x-2         x2-± 

\»  +  «/       y      \2/  x  +  yj 

(i-^L_+-^Yi+-i 2-Y 


19. 


20    a:2  -  ay  +  y2  ?(  s8  -  y8  ;   (y  -  x)2 
x2  -\-  xy  +  y2      x8  -\-  y8  '   (x  +  y)2 

2a  —  b  —  c  2b  —  c  —  a  2c  —  a  —  b 

(a  —  b)  (a  —  c)       (b  —  c)  (6  —  a)       (c  —  a)  (c  —  £) 

-J ? 1  +  1       1  +  1 

22<    s  +  1       (s  +  2)(g  +  l)  ^  g«^>      a rJ  ft 


a; +  2       (a  +  l)(a  +  2)  a2      £2      a 

V         1+*        1-x2  J\2x  +  lJ 

25>  a;8~8y3  x   ^2 - xy  +  y2   x  x (x* -  v*) . 

x  (x  —  y)       x2  +  2  xy  +  4:  y2         x8  +  y* 

no     2,2,2      b  +  c-a      c  +  a-b      a  +  b-c 

26.    — ttH ; 7 

a       b       c  be  ac  ab 


CHAPTER  X. 
FRACTIONAL    EQUATIONS. 


Reduction  of  Equations  Containing  Fractions. 

1.  Solve  -  —  =  x  —  9. 

o         11 

Multiply  by  33,  the  L.  C.  M.  of  the  denominators. 
Then,  llx  -  3x  +  3  =  33x  —  297, 

11  x  —  3x-33x  =  —  297  —  3, 
-25x  =  ~300. 
.-.  x  =  12. 

Note.  Since  the  minus  sign  precedes  the  second  fraction,  in 
removing  the  denominator  the  sign  of  every  term  of  the  numerator 
is  changed. 

2.  Solve 2x  +  1      2*-1  8 


2x-l      2cc  +  l      4a2-l 

The  L.  C.  D.  ==  (2x  +  1)  (2x  -  1). 
Multiply  by  the  L.  C.  D.,  and  we  have, 

4x2  +  4x  +  l-(4x2-4x  +  l)  =  8. 

.-.  4x2  +  4x  +  1-4x2  +  4x-  1  =  8. 

Reducing,  x  =  1. 

174.    To  Clear  an  Equation  of  Fractions,  therefore, 
Multiply  each  term  by  the  L.  C.  M.  of  the  denominators. 

If  a  fraction  is  preceded  by  a  minus  sign,  the  sign  of 
every  term  of  the  numerator  must  be  changed  when  the 
denominator  is  removed. 


Solve 


FRACTIONAL  EQUATIONS.  149 

x  —  4      x  —  5      a;  —  7      a;  —  8 
a;  —  5      a;  —  6      a;  —  8      a;  —  9 


Note.  The  solution  of  this  and  similar  problems  will  be  much 
easier  by  combining  the  fractions  on  the  left  side  and  the  fractions 
on  the  right  side  than  by  the  rule. 

(x  -  4)  (x  -  6)  -  (x  -  5)2  _  (x  -  7)  (x  -  9)  -  (x  -  8)2 
(x  -  5)  (x  -  6)  (x  —  8)  (x  —  9) 

By  simplifying  the  numerators,  we  have 
-1  -1 


(x  -  5)  (x  —  6)      (x  —  8)  (x  —  9) 

Since  the  numerators  are  equal,  the  denominators  are  equal. 

Hence,  (x  —  5)  (x  -  6)  =  (x  —  8)  (x  —  9). 

Solving,  we  have  x  =  7. 

Exercise  64. 
Solve: 
3«-l      2x  +  l      Ax-5       . 

h  — 3 5~  =  4* 

5x  +  l  ,  19cc  +  7      3z-l      7z-l 


3. 


4.    4  + 


3  9  2  6 

x      Sx-2      \\x  +  2      2-7x 


7  2  14 


9s +  5      8s-7_36a3  +  l5      41 
14      +      14       '         56        +56* 

7  a; -4      3a; -1      5(x- 1)  _3  (Sx  -  1)      x 
9  6  6        ~         20  7* 


150  FRACTIONAL  EQUATIONS. 

4  5  4  3 

-     10  a; +  11      12 a?  -13       .       7 -6a 

9 4  — . 

6  3  4 

3  5  9  1 

10'    y-4  +  2(y-4)"h2(2/-4)      2* 

2  5  8  x  5_ 

x-1      2  (x-1)      3(a?-l)      aj-l  +  18* 

2  a; -3      fi       a;  +  5        11 
2a; -4  3a; -6       2' 

10 -7a;      5a;-4        ,     2x  +  l  8  2a;-l 

13.    — = —  =  — = 17. 


6 -7a;  5x  2x-l      4a;2 -1      2x  + 1 

5  +  8x      45-Sx  5-2x      2-lx      5a;2  +  4 

14'    3  +  2a;~13-2a;'     1        x-1        x  +  1  ~  x2  -  1  ' 

15     5x-l_5x-S  _6 *_+?  .  _£f_  =  0 

2a;  +  3      2a;-3  a;  +  2      a;-2      aa-4 

a;      a;2  -5x      2  4       ,  a?  + 1      a2 -3 

16'   3~3^T~3'        20*    l+a;  +  l-a;      l-^-U* 

2a;  +  l    _7a;-l      2a;2-3a;-45 
3a;-3~~6a;  +  6  4a;2 -4 

rto     x2-a;  +  l   ,  a;2  +  a;  +  1       0 

22. H rr-= =  2a;. 

as— 1  a;  4-  1 

9a; +  5        3s2  -  51a;  -  71       15a;  -  7 
23'    G^-l)4"       18(a;2-l)  9(a;  +  l)' 

24;    a7+^  +  a7+~3~a;2  +  5a;  +  6==a 

25    _1 1_  =  _1 1_. 

x  —  1       a;  —  2      a;  —  3      a;  —  4 


FRACTIONAL  EQUATIONS.  151 

175.  If  the  denominators  contain  both  simple  and  com- 
pound expressions,  it  is  best  to  remove  the  simple  expres- 
sions first,  and  then  each  compound  expression  in  turn. 
After  each  multiplication  the  result  should  be  reduced  to 
the  simplest  form. 

i.   solve  -ir-  +  5_jrg  =  _r-. 
Multiply  both  sides  by  14, 

Then,  8x  +  5  +  7('7X~-^-)  =  8x  +  12. 

'  3a:  4-  1 

Transpose  and  combine,    — ^ — ;r— p  =  7. 

Divide  by  7  and  multiply  by  3x  +  1, 

7x  — 3  =  3z  +  l. 
i 
.-.  x  =  1- 

Q      4a;  7x 

9  Q1     3~T     l     T"3 

2.    Solve  _j_  =  z jjp- 

Simplify  the  complex  fractions  by  multiplying  both  terms  of  each 
fraction  by  9. 


Solve : 


27 -4x 

1 

7x-27 

36 

4 

90 

Multiply  both  sides  by  180, 

the  L.  C.  D., 

135- 20  x 

=  45 

-  14  x  +  54, 

—  6x 

=  - 

36. 

.*.  X 

=  6. 

Exercise  65 

e : 

1. 

4a;  +  3      2x- 

5 

2a;-l 

10           5x- 

1 

5 

2* 

9a;  +  20      4a;- 
36            5x 

-12 
-4 

*i; 

152  FRACTIONAL  EQUATIONS. 

10a;  +  17       12a; +  2       5a; -4 
3'  18  13  oj  -16"        9 

6a;  +  7      7a;-13      2a;  +  4 

9       +  6z  +  3  "       3 
6a;  +  7      2a;-22a;  +  l 

15  7a;-6~~       5 

6a;  +  l        2a;-4       2x-l 

15  7a;-16~       5 

11  a; -13      22  a; -75        13  x  +  7 


5. 


7. 


9, 


14  28  2(3a  +  7) 

2a;  +  8^       13a;-2       x      Ix      x  + 16 
9  17  x  -  32  +  3  7  12  36 


5a;        7-2a;2        l  +  3a;      2 


./• 


^+4, 


15  14  (a; -1)  21  6        ^105 


10    2*-5,      *~3        4*~3      iy 
10*         5       +2a;-15~      10  1A" 


Literal  Equations. 

176.    Literal  equations  are  equations  in  which  some  or 
all  of  the  given  numbers  are  represented  by  letters ;  known 
numbers  are  represented  by  the  first  letters  of  the  alphabet- 
Solve  {a  -  x)  (a  +  x)  =  2  a2  +  2aa;  -  x\ 

Then,  a2  —  x2  =  2  a2  +  2  ax  —  x2, 

—  2  ax  =  a2, 
a 

Exercise  66. 
Solve : 

1.  ax  +  2  J  =  3  bx  +  4  a.     3.  (a  +  a;)  ( b  +  x)  —  x  (x  —  c). 

2.  x2  +  &2  =  (a-a;)2.  4.   (x- a)(x  +  b)=(x-b)(x- c). 


ERA C TIONAL  EQUA  TIONS. 


153 


8. 
9 
.0. 

Ll. 

12. 
13. 

14. 


a  +  ax      c  -\-  ex 

C  +  d  Til  —  x 

ab  -f  bx      an  ■+-  nx 

x  -\-  2  _m  -{-  n 
x  —  2      m  —  n 

m  -\-  n      m  —  n 


2-x 
e  -\-  dx 


2  +  x 

a  -\-  bx  _ 
a  -\-  b        e  -\-  d 

6x  +  a  _3x  — 
±x  +  b 


X         X         X 

a      b       c 


ax  —  b 


2x  —  a 
=  d. 
bx  +  o 


e 

5  ax 

a  —  b 

x 


=  abc. 


3a  =  8#. 

5  a  2bx 


a  — 
I    15.    -  + 


b      a  +  b      a2 
n  x  +  n 


16. 


x-\-n 


x 


nx 
a 


a      b  —  a      b  +  a 

x  +  a 


17. 


18. 


19. 


20. 


21. 


22. 


23. 


24. 


ic  —  a 


x  —  b 

\2a;-  6 

x  —  a 

2 

2x  —  a 

m  -\-  n 

X 

a 

1 

b 

ax 


b      ax  +  b 


ex  +  d 


ex 
1 

a  ■+•  x 


a  —  x  ■+-  3 
3 

a  —  b 
bx  -+-  c 


2rf 
a 


+ 


a#  —  c 


0. 


25. 


x  =  b 


-b    _2(a-\-b) 
x  —  a      x-\-b  x 

)   .   c  —  bx 


20  a  —  bx   ,  9  e  —  asc   , 

26.    - H s 1- 


6 
6d  —  c# 


5a 


27.    --       &-s   ,  a(6-s) 


ax 
T 


2c 


Se 

b_ 
Sd 


2d 


=  10. 


154  FRACTIONAL  EQUATIONS. 

Problems  Involving  Fractional  Equations. 
Exercise  67. 

1.  The  sum  of  the  third  and  fourth  parts  of  a  certain 
number  exceeds  3  times  the  difference  of  the  fifth  and  sixth 
parts  by  29.     Find  the  number. 

Let  x  s=  the  number. 

Then,    -  +  -  =  the  sum  of  its  third  and  fourth  parts, 

x      x 

t  —  -  =  the  difference  of  its  fifth  and  sixth  parts, 

3^-  —  -J=3  times  the  difference  of  its  fifth  and  sixth  parts, 

x      x      0/x      x\  . 

o  +  T  —  of-  —  -J  =  the  given  excess. 

But  29  =  the  given  excess. 

•••i+i-(f-i)=* 

Multiply  by  60,  the  L.  C.  D.  of  the  fractions. 

20s  +  15x  —  36x  +  SOx  =  60  X  29. 
Combining,  29  x  =  60  X  29. 

Dividing  by  29,  x  =  60. 

The  required  number,  therefore,  is  60. 

2.  The  difference  between  the  fifth  and  ninth  parts  of 
a  certain  number  is  4.     Find  the  number. 

3.  One  half  of  a  certain  number  exceeds  the  sum  of  its 
fifth  and  ninth  parts  by  17.     Find  the  number. 

4.  The  sum  of  the  third  and  sixth  parts  of  a  certain 
number  exceeds  the  difference  of  its  sixth  and  seventh  parts 
by  20.     Find  the  number. 

5.  There  are  two  consecutive  numbers,  x  and  x  +  1,  such 
that  one  half  of  the  larger  exceeds  one  third  of  the  smaller 
number  by  9.     Find  the  numbers. 


FRACTIONAL  EQUATIONS.  155 

6.  The  sum  of  two  numbers  is  63,  and  if  the  greater 
is  divided  by  the  smaller  number,  the  quotient  is  2  and  the 
remainder  3.     Find  the  numbers. 

Let  x  =  the  greater  number. 

Then  63  —  x  =  the  smaller  number. 

Dividend  —  Remainder 


Since  the  quotient  = 


Divisor 


and  since,  in  this  problem,  the  dividend  is  x,  the  remainder  is  3,  and 
the  divisor  is  63  —  x,  we  have 


63  -x 
Solving,  x  =  43. 

The  two  numbers,  therefore,  are  43  and  20. 

7.  The  sum  of  two  numbers  is  120,  and  if  the  greater  is 
divided  by  the  smaller  number,  the  quotient  is  2  and  the 
remainder  15.     Find  the  numbers. 

8.  The  sum  of  two  numbers  is  126,  and  if  the  greater  is 
divided  by  the  smaller  number,  the  quotient  is  3  and  the 
remainder  10.     Find  the  numbers. 

9.  The  difference  of  two  numbers  is  51,  and  if  the  greater 
is  divided  by  the  smaller,  the  quotient  is  4  and  the  remain- 
der 6.     Find  the  numbers. 

10.  The  difference  of  two  numbers  is  87,  and  if  the 
greater  is  divided  by  the  smaller,  the  quotient  is  8  and  the 
remainder  10.     Find  the  numbers. 

11.  Divide  450  into  two  parts  such  that  the  smaller  part 
is  contained  in  the  larger  part  9  times,  with  a  remainder 
of  20. 

12.  The  difference  of  two  numbers  is  25,  and  if  the 
greater  is  divided  by  the  smaller,  the  quotient  is  4  and 
the  remainder  4.     Find  the  numbers. 


156  FRACTIONAL  EQUATIONS. 

13.  Eight  years  ago  a  boy  was  one  fourth  as  old  as  he 
will  be  one  year  hence.     How  old  is  he  now  ? 

Let  x  =  the  number  of  years  old  he  is  now. 

Then  x  —  8  =  the  number  of  years  old  he  was  eight  years  ago, 
and         x  +  1  =  the  number  of  years  old  he  will  be  one  year  hence. 
.\x  —  S  =  i(x  +  1). 
Solving,  x  —  11. 

Therefore,  the  boy  is  11  years  old. 

14.  A  son  is  one  third  as  old  as  his  father.     In  10  years 
he  will  be  one  half  as  old.     Find  the  age  of  the  son. 

15.  B's  age  is  one  fifth  of  A's  age.     In  12  years  B's  age 
will  be  one  third  of  A's  age.     Find  their  ages. 

16.  The  sum  of  the  ages  of  A  and  B  is  60  years,  and  10 
years  hence  B's  age  will  be  one  third  of  A's.     Find  their 


17.  A  father  is  40  years  old,  and  his  son  is  one  fourth  of 
that  age.  In  how  many  years  will  the  son  be  half  as  old 
as  his  father  ? 

18.  A  is  30  years  old,  and  B's  age  is  two  thirds  of  A's. 
How  many  years  ago  was  B's  age  one  third  of  A's  ? 

19.  A  son  is  one  fourth  as  old  as  his  father.  Four  years 
ago  he  was  only  one  fifth  as  old  as  his  father.  What  is 
the  age  of  each  ? 

20.  A  is  40  years  old,  and  B  is  half  as  old  as  A.  In  how 
many  years  will  B  be  two  thirds  as  old  as  A  ? 

21.  B  is  one  third  as  old  as  A.  Ten  years  ago  he  was 
one  fourth  as  old  as  A.     What  are  their  present  ages  ? 

22.  The  sum  of  the  ages  of  a  father  and  his  son  is  75 
years.  The  son's  age  increased  by  5  years  is  one  fourth  of 
the  father's  age.     Find  their  ages. 


FRACTIONAL  EQUATIONS.  157 

23.  A  rectangle  has  its  length  6  feet  more  and  its  width 
5  feet  less  than  the  side  of  the  equivalent  square.  Find  the 
dimensions  of  the  rectangle. 

Let  x  =  the  number  of  feet  in  a  side  of  the  square. 

Then  x  +  6  =  the  number  of  feet  in  the  length  of  the  rectangle, 
and        x  —  5  =  the  number  of  feet  in  the  width  of  the  rectangle. 

Since  the  area  of  a  rectangle  is  equal  to  the  product  of  the  number 
of  units  of  length  in  the  length  and  width  of  the  rectangle, 

(x  +  6)  (x  —  5)  =  the  area  of  the  rectangle  in  square  feet, 
and  x  x  x  =  the  area  of  the  square  in  square  feet. 

But  these  areas  are  equal. 

.-.  (x  +  6)  (x  -  5)  =  x2. 
Solving,  x  =  30. 

Therefore,  the  dimensions  of  the  rectangle  are  36  feet 
and  25  feet. 

24.  A  rectangle  has  its  length  and  breadth,  respectively, 
12  feet  longer  and  8  feet  shorter  than  the  side  of  the 
equivalent  square.     Find  its  area. 

25.  The  length  of  a  floor  exceeds  the  breadth  by  6  feet. 
If  each  dimension  were  1  foot  more,  the  area  of  the  floor 
would  be  41  sq.  ft.  more.     Find  its  dimensions. 

26.  A  rectangle  whose  length  is  8  feet  more  than  its 
breadth  would  have  its  area  35  sq.  ft.  more,  if  each  dimen- 
sion were  1  foot  more.     Find  its  dimensions. 

27.  The  length  of  a  rectangle  exceeds  its  width  by  4  feet. 
If  the  length  were  diminished  by  2  feet  and  the  width  by 
2  feet,  the  area  would  be  diminished  by  40  sq.  ft.  Find  its 
dimensions. 

28.  The  length  of  a  floor  exceeds  its  width  by  8  feet. 
If  each  dimension  were  2  feet  more,  the  area  would  be  124 
sq.  ft.  more.     Find  its  dimensions. 


158  FRACTIONAL  EQUATIONS. 

29.  A  can  do  a  piece  of  work  in  2  days,  and  B  can  do 
it  in  3  days.  How  many  days  will  it  take  both  together  to 
do  the  work  ? 

Let        x  ~  the  number  of  days  it  will  take  both  together. 

Then     -  =  the  part  both  together  can  do  in  one  day, 

x 

■£•  =  the  part  A  can  do  in  one  day, 

£  s=  the  part  B  can  do  in  one  day, 

and     i  +  i  =  the  part  both  together  can  do  in  one  day. 

,.1+1=1. 

2      3      x 

Solving,  x  =  1£. 

Therefore,  they  together  can  do  the  work  in  1^  days. 

30.  A  can  do  a  piece  of  work  in  3  days,  B  in  4  days,  and 
C  in  5  days.  How  many  days  will  it  take  them  to  do  it 
working  together  ? 

31.  A  can  do  a  piece  of  work  in  6  days,  B  in  5  days,  and 
C  in  4  days.  How  many  days  will  it  take  them  together 
to  do  the  work  ? 

32.  A  can  do  a  piece  of  work  in  2\  days,  B  in  3^  days, 
and  C  in  3|  days.  How  many  days  will  it  take  them 
together  to  do  the  work  ? 

33.  A  can  do  a  piece  of  work  in  8  days,  B  in  10  days ; 
A  and  B  together,  with  the  help  of  C,  can  do  the  work  in 
3  days.  How  many  days  will  it  take  C  alone  to  do  the 
work? 

.34.  A  and  B  together  can  mow  a  field  in  8  honrs,  A  and 
C  in  10  hours,  and  A  alone  in  15  hours.  In  how  many 
hours  can  B  and  C  together  mow  the  field  ?  . 

!  35.  A  and  B  together  can  build  a  wall  in  12  days,  A  and 
C  in  15  days,  B  and  C  in  20  days.  In  how  many  days  can 
they  build  the  wall  if  they  all  work  together  ? 

Hint.     By  working  2  days  each  they  build  y1^  +  ^  4-  ^  of  it. 

Hence,  m  one  day  they  can  build  ±(fa  +  TV  +  **&)• 


FRA C TIONAL  EQUA TIONS.  159 

36.  A  cistern  can  be  filled  by  two  pipes  in  15  and  20 
hours,  respectively ;  and  can  be  emptied  by  a  waste  pipe 
in  30  hours.  In  how  many  hours  will  it  be  filled  if  all  the 
pipes  together  are  open  ? 

Let  x  =  the  number  of  hours  it  all  the  pipes  are  open. 

Then  -  =  the  part  filled  in  one  hour  if  all  the  pipes  are  open. 

tV  +  -fa  ~~  ^5  =  the  part  all  together  can  fill  in  one  hour. 

*'*15+20      30      x 

Solving,  x  =3  12. 

Therefore,  if  all  the  pipes  are  open  it  will  be  filled  in  12 
hour?. 

37.  A  cistern  can  be  filled  by  three  pipes  in  8,  12,  and 
16  hours,  respectively.  In  how  many  hours  will  it  be  filled 
by  all  the  pipes  together  ? 

38.  A  cistern  can  be  filled  by  two  pipes  in  4  hours  and  5 
hours,  respectively,  and  can  be  emptied  by  a  third  pipe  in 
6  hours.  In  how  many  hours  will  the  cistern  be  filled  if  the 
pipes  are  all  running  together  ? 

39.  A  tank  can  be  filled  by  three  pipes  in  1  hour  and  40 
minutes,  3  hours  and  20  minutes,  and  5  hours,  respectively. 
In  what  time  will  the  tank  be  filled  if  all  three  pipes  are 
running  together  ? 

40.  A  cistern  can  be  filled  by  three  pipes  in  2\  hours,  Z\ 
hours,  and  4§  hours,  respectively.  In  how  many  hours  will 
the  cistern  be  filled  if  all  the  pipes  are  running  together  ? 

41.  A  cistern  has  three  pipes.  The  first  pipe  will  fill  the 
cistern  in  6  hours,  the  second  in  10  hours,  and  all  three 
pipes  together  will  fill  it  in  3  hours.  How  long  will  it  take 
the  third  pipe  alone  to  fill  it  ? 


160  FRACTIONAL  EQUATIONS. 

42.  A  courier  who  travels  6  miles  an  hour  is  followed, 
after  2  hours,  by  a  second  courier  who  travels  7£  miles  an 
hour.  In  how  many  hours  will  the  second  courier  overtake 
the  first  ? 

Let  x  =  the  number  of  hours  the  first  travels. 

Then  x  —  2  =  the  number  of  hours  the  second  travels, 

6  x  =  the  number  of  miles  the  first  travels, 
and  (x  —  2)  7£  =  the  number  of  miles  the  second  travels. 

They  both  travel  the  same  distance. 

.:6x=  (x  — 2)7*, 
or  12x=15x-30. 

.-.  x  =  10. 

Therefore,  the  second  courier  will  overtake  the  first  in 
10  —  2,  or  8  hours. 

43.  A  sets  out  from  Boston  and  walks  towards  Portland 
at  the  rate  of  3  miles  an  hour.  Three  hours  afterward  B 
sets  out  from  the  same  place  and  walks  in  the  same  direc- 
tion at  the  rate  of  3^  miles  an  hour.  How  far  from  Boston 
will  B  overtake  A  ? 

44.  A  courier  who  goes  at  the  rate  of  4 \  miles  an  hour  is 
followed,  after  4  hours,  by  another  who  goes  at  the  rate  of 
5*  miles  an  hour.  In  how  many  hours  will  the  second 
overtake  the  first  ? 

45.  A  person  walks  to  the  top  of  a  mountain  at  the  rate 
of  1^  miles  an  hour,  and  down  the  same  way  at  the  rate  of 
4*  miles  an  hour.  If  he  is  out  6  hours,  how  far  is  it  to  the 
top  of  the  mountain  ? 

46.  In  going  a  certain  distance,  a  train  traveling  at  the 
rate  of  50  miles  an  hour  takes  2  hours  less  than  a  train 
traveling  40  miles  an  hour.     Find  the  distance. 


FRACTIONAL  EQUATIONS.  161 

47.  Find  the  time  between  2  and  3  o'clock  when  the 
hands  of  a  clock  are  together. 

At  2  o'clock  the  hour-hand  is  10  minute-spaces  ahead  of  the 
minute-hand. 

Let  x  =  the  number  of  spaces  the  minute-hand  moves  over. 

Then  x  —  10  =  the  number  of  spaces  the  hour-hand  moves  over. 
Now,  as  the  minute-hand  moves  12  times  as  fast  as  the  hour-hand, 
12  (x  —  10)  =  the  number  of  spaces  the  minute-hand  moves  over. 

.-.  12  (x  -  10)  =  x, 
and  11  x  =  120. 

.-.  x  =  lOf-f 

Therefore,  the  time  is  10}£  minutes  past  2  o'clock. 

48.  Find  the  time  between  4  and  5  o'clock  when   the 
hands  of  a  clock  are  together. 

49.  Find  the  time  between  3  and  4  o'clock  when  the 
hands  of  a  clock  are  at  right  angles  to  each  other. 

Hint.     In  this  case  the  minute-hand  is  16  minutes  ahead  of  the 
hour-hand. 

50.  Find  the  time  between  2  and  3  o'clock  when  the 
hands  of  a  clock  point  in  opposite  directions. 

Hint.     In  this  case  the  minute-hand  is  30  minutes  ahead  of  the 
hour-hand,  or  30  minutes  behind  it. 

51.  Find  the  times  between  4  and  5  o'clock  when  the 
hands  of  a  clock  are  at  right  angles  to  each  other. 

52.  Find  the  time  between  1  and  2  o'clock  when  the 
hands  of  a  clock  point  in  opposite  directions. 

53.  At  what  time  between  6  and  7  o'clock  are  the  hands 
of  a  watch  together  ? 


162  FRACTIONAL  EQUATIONS. 

54.  A  hare  takes  4  leaps  to  a  greyhound's  3;  but  2  of 
the  greyhound's  leaps  are  equivalent  to  3  of  the  hare's. 
The  hare  has  a  start  of  50  of  her  own  leaps.  How  many 
leaps  must  the  greyhound  take  to  catch  the  hare  ? 

Let  3  x  =  the  number  of  leaps  taken  by  the  greyhound. 

Then       4  x  =  the  number  of  leaps  of  the  hare  in  the  same  time. 
Also,  let   a  =  the  number  of  feet  in  one  leap  of  the  hare. 

Then      —  =  the  number  of  feet  in  one  leap  of  the  hound. 

Therefore,        3  x  X  — -  or  — —  =  the  whole  distance. 

As  the  hare  has  a  start  of  50  leaps,  and  takes  4  x  leaps  more  before 

she  is  caught,  and  as  each  leap  is  a  feet, 

(50  +  4  x)  a  =  the  whole  distance. 

9  ax 
.-.  -7r-=  (50  +  4x)a. 

Multiply  by  2,  9  ax  =  (100  +  8  x)  a. 

Divide  by  a,  9x "=  100  +  8x, 

x  ==  100, 
.'.Sx  =  300. 

Therefore,  the  greyhound  must  take  300  leaps. 

55.  A  hound  takes  3  leaps  while  a  rabbit  takes  5  j  but 
1  of  the  hound's  leaps  is  equivalent  to  2  of  the  rabbit's. 
The  rabbit  has  a  start  of  120  of  her  own  leaps.  How 
many  leaps  will  the  rabbit  take  before  she  is  caught  ? 

56.  A  rabbit  takes  6  leaps  to  a  dog's  5,  and  7  of  the  dog's 
leaps  are  equivalent  to  9  of  the  rabbit's.  The  rabbit  has  a 
start  of  60  of  her  own  leaps.  How  many  leaps  must  the 
dog  take  to  catch  the  rabbit  ? 

57.  A  dog  takes  4  leaps  while  a  rabbit  takes  5  j  but  3 
of  the  dog's  leaps  are  equivalent  to  4  of  the  rabbit's.  The 
rabbit  has  a  start  of  90  of  the  dog's  leaps.  How  many 
leaps  will  each  take  before  the  rabbit  is  caught  ? 


FRACTIONAL  EQUATIONS.  163 

58.  A  merchant  adds  yearly  to  his  capital  one  third  of 
it,  but  takes  from  it,  at  the  end  of  each  year,  $  5000  for 
expenses.  At  the  end  of  the  third  year,  after  deducting 
the  last  $5000,  he  has  twice  his  original  capital.  How 
much  had  he  at  first? 

Let  x  =  number  of  dollars  he  had  at  first. 

4x      ^™       4  a; -15000 
Then        -g-  —  5000,  or > 

will  stand  for  the  number  of  dollars  at  the  end  of  first  year, 
4  /4«  -  15000 \       ennn        16 x  -  105000 

"^  s( 8 )  ~  5°00'  °r 9 ' 

will  stand  for  the  number  of  dollars  at  the  end  of  second  year, 
4  /16s-  105000 \       CAAA        64^-555000 

ftnd  8  ( 0 )  ""  500°'  °r  _ "27 ' 

will  stand  for  the  number  of  dollars  at  the  end  of  third  year. 

But  2  x  stands  for  the  number  of  dollars  at  the  end  of  third  year. 
64  a; -555000      . 

'•• 27 =  2X' 

Whence  x  =  55,500. 

Therefore,  the  merchant  had  $55,500  at  first. 

59.  A  trader  adds  yearly  to  his  capital  one  fourth  of  it, 
but  takes  from  it,  at  the  end  of  each  year,  $800  for  ex- 
penses. At  the  end  of  the  third  year,  after  deducting  the 
last  $800,  he  has  1$  of  his  original  capital.  How  much 
had  he  at  first  ? 

60.  A  trader  adds  yearly  to  his  capital  one  fifth  of  it, 
but  takes  from  it,  at  the  end  of  each  year,  $2500  for  ex- 
penses. At  the  end  of  the  third  year,  after  deducting 
the  last  $2500,  he  has  1T7^  of  his  original  capital.  Find 
his  original  capital. 

61.  A  trader  maintained  himself  for  three  years  at  an  ex- 
pense of  $500  a  year;  and  each  year  increased  that  part  of 
his  stock  which  was  not  so  expended  by  one  third  of  it. 
At  the  end  of  the  third  year  his  original  stock  was  doubled. 
What  was  his  original  stock  ? 


164  FRACTIONAL  EQUATIONS. 

62.  The  sum  of  the  third,  fourth,  and  fifth  parts  of  a 
number  exceeds  the  half  of  the  number  by  17.  Find  the 
number. 

63.  There  are  two  consecutive  numbers,  x  and  a?  -f  1, 
such  that  one  fourth  of  the  smaller  exceeds  one  ninth  of  the 
larger  by  11.     Find  the  numbers. 

64.  Find  three  consecutive  numbers  such  that  if  they  are 
divided  by  7,  10,  and  17,  respectively,  the  sum  of  the  quo- 
tients will  be  15. 

65.  In  a  mixture  of  alcohol  and  water  the  alcohol  was 
24  gallons  more  than  half  the  mixture,  and  the  water  was 
4  gallons  less  than  one  fourth  the  mixture.  How  many 
gallons  were  there  of  each  ? 

66.  The  width  of  a  room  is  three  fourths  of  its  length. 
If  the  width  was  4  feet  more  and  the  length  4  feet  less,  the 
room  would  be  square.     Find  its  dimensions. 

67.  The  difference  of  two  numbers  is  40,  and  if  the  greater 
is  divided  by  the  less  the  quotient  is  4,  and  the  remainder 
4.     Find  the  numbers. 

68.  Divide  the  number  240  into  two  parts  such  that  the 
smaller  part  is  contained  in  the  larger  part  5  times,  with  a 
remainder  of  6. 

69.  A  can  do  a  piece  of  work  in  3  days,  B  in  4  days,  and 
C  in  6  days.  How  many  days  will  it  take  them  to  do  it 
working  together  ? 

70.  At  what  time  between  2  and  3  o'clock  are  the  hands 
of  a  watch  at  right  angles  ? 

71.  Find  a  number  such  that  the  sum  of  its  sixth  and 
ninth  parts  shall  exceed  the  difference  of  its  ninth  and 
twelfth  parts  by  9. 


FRACTIONAL  EQUATIONS.  165 

72.  The  sum  of  two  numbers  is  91,  and  if  the  greater  is 
divided  by  the  less  the  quotient  is  2,  and  the  remainder 

I  is  1.     Find  the  numbers. 

73.  A  is  60  years  old,  and  B  is  three  fourths  as  old. 
How  many  years  since  B  was  just  half  as  old  as  A  ? 

74.  A  can  do  a  piece  of  work  in  2£  days,  B  in  3£  days, 
and  C  in  4§  days.  How  long  will  it  take  them  to  do  it 
working  together  ? 

75.  A  and  B  can  separately  do  a  piece  of  work  in  12 
days  and  20  days,  and  with  the  help  of  C  they  can  do  it  in 
6  days.     How  long  would  it  take  C  alone  to  do  the  work  ? 

76.  A  rectangle  has  its  length  4  feet  longer  and  its  width 
3  feet  shorter  than  the  side  of  the  equivalent  square.  Find 
its  area. 

77.  A  person  has  6  hours  at  his  disposal.  How  far  may 
he  ride  at  6  miles  an  hour  so  as  to  return  in  time,  walking 

j  'back  at  the  rate  of  3  miles  an  hour  ? 

78.  A  boy  starts  from  Exeter  and  walks  towards  Ando- 
ver  at  the  rate  of  3  miles  an  hour,  and  2  hours  later  another 
boy  starts  from  Andover  and  walks  towards  Exeter  at  the 

*  rate  of  2\  miles  an  hour.     The  distance  from  Exeter  to 

tA  ndover  is  28  miles.  How  far  from  Exeter  will  they  meet  ? 
79.  A  dog  makes  4  leaps  while  a  hare  makes  5,  but  3  of 
e  dog's  leaps  are  equal  to  4  of  the  hare's.  The  hare  has  a 
start  of  60  of  the  dog's  leaps.  How  many  leaps  will  each 
make  before  the  hare  is  caught  ? 

80.  At  what  time  between  3  and  4  o'clock  are  the  hands 
f  a  watch  pointing  in  opposite  directions  ? 

81.  In  going  from  Boston  to  Portland,  a  passenger  train, 
|  36  miles  an  hour,  occupies  1  hour  less  time  than  a  freight 

in  at  27  miles  an  hour.     Find  the  distance  from  Boston 
Portland. 


166  FRACTIONAL  EQUATIONS. 

82.  A  cistern  can  be  filled  by  three  pipes  in  15,  20,  and 
30  minutes,  respectively.  In  what  time  will  it  be  filled  if 
the  pipes  are  all  running  together  ? 

83.  A  cistern  can  be  filled  by  two  pipes  in  25  minutes 
and  30  minutes,  respectively,  and  emptied  by  a  third  in  20 
minutes.  In  what  time  will  it  be  filled  if  all  three  pipes 
are  running  together  ? 

84.  A  hare  takes  7  leaps  while  a  dog  takes  5,  and  5  of 
the  dog's  leaps  are  equal  to  8  of  the  hare's.  The  hare  has 
a  start  of  50  of  her  own  leaps.  How  many  leaps  will  the 
hare  take  before  she  is  caught  ? 

85.  The  width  of  a  rectangle  is  an  inch  more  than  half 
its  length,  and  if  a  strip  5  inches  wide  is  taken  off  from  the 
four  sides,  the  area  of  the  strip  is  510  square  inches.  Find 
the  dimensions  of  the  rectangle. 

86.  A  and  B  together  can  do  a  piece  of  work  in  10  days, 
A  and  C  in  12  days,  and  A  by  himself  in  18  days.  How 
many  days  will  it  take  B  and  C  together  to  do  the  work  ? 
How  many  days  will  it  take  A,  B,  and  C  together  ? 

87.  A  and  B  can  do  a  piece  of  work  in  10  days,  A  and 
G  in  12  days,  B  and  C  in  15  days.  How  long  will  it  take 
them  to  do  the  work  if  they  all  work  together  ? 

88.  A  sets  out  and  travels  at  the  rate  of  9  miles  in  2 
hours.  Seven  hours  afterwards  B  sets  out  from  the  same 
place  and  travels  in  the  same  direction  at  the  rate  of  5 
miles  an  hour.     In  how  many  hours  will  B  overtake  A  ? 

89.  A  man  walks  to  the  top  of  a  mountain  at  the  rate  of 
2\  miles  an  hour,  and  down  the  same  way  at  the  rate  of  4 
miles  an  hour,  and  is  out  5  hours.  How  far  is  it  to  the  top 
of  the  mountain  ? 


FRACTIONAL  EQUATIONS.  167 

90.  A  tank  can  be  filled  by  three  pipes  in  1  hour  20 
minutes,  2  hours  20  minutes,  and  3  hours  20  minutes, 
respectively.  In  how  many  minutes  can  it  be  filled  by 
all  three  together? 

91.  A's  age  now  is  two  fifths  of  B's.  Eight  years  ago 
A's  age  was  two  ninths  of  B's.     Find  their  ages. 

92.  A  had  five  times  as  much  money  as  B.  He  gave  B 
5  dollars,  and  then  had  only  twice  as  much  as  B.  How 
much  had  each  at  first  ? 

93.  At  what  time  between  12  and  1  o'clock  are  the  hour 
and  minute-hands  pointing  in  opposite  directions  ? 

94.  Eleven  sixteenths  of  a  certain  principal  was  at  in- 
terest at  5  per  cent,  and  the  balance  at  4  per  cent.  The 
entire  income  was  $1500.     Find  the  principal. 

95.  A  train  that  travels  36  miles  an  hour  is  f  of  an 
hour  in  advance  of  a  second  train  that  travels  42  miles 
an  hour.  In  how  long  a  time  will  the  last  train  overtake 
the  first? 

96.  An  express  train  that  travels  40  miles  an  hour  starts 
from  a  certain  place  50  minutes  after  a  freight  train,  and 
overtakes  the  freight  train  in  2  hours  5  minutes.  Find 
the  rate  per  hour  of  the  freight  train. 

97.  If  1  pound  of  tin  loses  /T  of  a  pound,  and  1  pound 
of  lead  loses  -fa  of  a  pound,  when  weighed  in  water,  how 
many  pounds  of  tin  and  of  lead  in  a  mass  of  60  pounds  that 
loses  7  pounds  when  weighed  in  water  ? 

98.  If  19  ounces  of  gold  lose  1  ounce,  and  10  ounces  of 
silver  lose  1  ounce,  when  weighed  in  water,  how  many 
ounces  of  gold  and  of  silver  in  a  mass  of  gold  and  silver 
that  weighs  530  ounces  in  air  and  495  ounces  in  water  ? 


168  FRACTIONAL  EQUATIONS. 

99.  A  messenger  starts  to  carry  a  despatch,  and  5  hours 
later  a  second  messenger  sets  out  to  overtake  the  first  in  8 
hours.  In  order  to  do  this,  he  is  obliged  to  travel  2\  miles 
an  hour  more  than  the  first.  How  many  miles  an  hour- 
does  the  first  travel  ? 

100.  The  fore  and  hind  wheels  of  a  carriage  are  respec- 
tively 9^  feet  and  llf  feet  in  circumference.  What  distance 
will  the  carriage  have  made  when  one  of  the  fore  wheels 
has  made  160  revolutions  more  than  one  of  the  hind 
wheels  ? 

101.  When  a  certain  brigade  of  troops  is  formed  in  a 
solid  square  there  is  found  to  be  100  men  over ;  but  when 
formed  in  column  with  5  men  more  in  front  and  3  men  less 
in  depth  than  before,  the  column  needs  5  men  to  complete 
it.     Find  the  number  of  troops. 

102.  An  officer  can  form  his  men  in  a  hollow  square  14. 
deep.  The  whole  number  of  men  is  3136.  Find  the  num- 
ber of  men  in  the  front  of  the  hollow  square. 

103.  A  trader  increases  his  capital  each  year  by  one 
fourth  of  it,  and  at  the  end  of  each  year  takes  out  $2400 
for  expenses.  At  the  end  of  3  years,  after  deducting  the 
last  $2400,  he  finds  his  capital  to  be  $10,000.  Find  his 
original  capital. 

104.  A  and  B  together  can  do  a  piece  of  work  in  1^  days, 
A  and  C  together  in  If  days,  and  B  and  C  together  in  If 
days.  How  many  days  will  it  take  each  alone  to  do  the 
work? 

105.  A  fox  pursued  by  a  hound  has  a  start  of  100  of  her 
leaps.  The  fox  makes  3  leaps  while  the  hound  makes  2; 
but  3  leaps  of  the  hound  are  equivalent  to  5  of  the  fox 
How  many  leaps  will  each  take  before  the  hound  catches 
the  fox? 


FRACTIONAL  EQUATIONS.  169 

Formulas  and  Rules. 

177.  When  the  given  numbers  of  a  problem  are  repre- 
sented by  letters,  the  result  obtained  from  solving  the  prob- 
lem is  a  general  expression  which  includes  all  problems  of 
that  class.  Such  an  expression  is  called  a  formula,  and  the 
translation  of  this  formula  into  words  is  called  a  rule. 

We  will  illustrate  by  examples: 

1.   The  sum  of  two  numbers  is  s,  and  their  difference  d) 


find  the 

numbers. 

Let 

x  =  the  smaller  number ; 

then, 

x  +  d  =  the  larger  number. 

Hence, 

x  +  x  +  d  =  a, 

or 

2x  —  s  —  d. 
s  —  d 

2 
s  +  d 


Therefore,  the  numbers  are  — - —  and  — - — 

As  these  formulas  hold  true  whatever  numbers  s  and  d 
stand  for,  we  have  the  general  rule  for  finding  two  numbers 
when  their  sum  and  difference  are  given : 

Add  the  difference  to  the  sum  and  take  half  the  result  for 
the  greater  number. 

Subtract  the  difference  from  the  sum  and  take  half  the 
result  for  the  smaller  number. 

2.  If  A  can  do  a  piece  of  work  in  a  days,  and  B  can  do 
the  same  work  in  b  days,  in  how  many  days  can  botb 
together  do  it? 


170  FRACTIONAL  EQUATIONS. 


Let  x  =  the  required  number  of  days. 

Then,  -  =  the  part  both  together  can  do  in  one  day. 

x 

Now,  -  =  the  part  A  can  do  in  one  day, 

and  -  as  the  part  B  can  do  in  one  day ; 

therefore,      -  +  -  =  the  part  both  together  can  do  in  one  day. 

,.!  +  i  =  i. 

a      b      x 


Whence, 


ab 


a  +  b 


The  translation  of  this  formula  gives  the  following  rule 
for  finding  the  time  required  by  two  agents  together  to 
produce  a  given  result,  when  the  time  required  by  each 
agent  separately  is  known : 

Divide  the  product  of  the  numbers  that  express  the  units 
of  time  required  by  each  to  do  the  work  by  the  sum  of  these 
numbers  ;  the  quotient  is  the  time  required  by  both  together. 

Exercise  68. 

1.  A  person  has  a  hours  at  his  disposal.  How  far  may 
he  ride  in  a  coach  that  travels  b  miles  an  hour,  so  as  to 
return  home  in  time,  walking  back  at  the  rate  of  c  miles  aa 
hour  ? 

2.  A  courier  who  travels  a  miles  a  day  is  followed  after 
c  days  by  another  who  travels  b  miles  a  day.  In  how  many 
days  will  the  second  overtake  the  first  ? 

3.  A  has  a  dollars  and  B  has  b  dollars.  B  gives  A  a  cer- 
tain number  of  dollars,  and  then  has  c  times  as  much  as  A 
How  many  dollars  does  A  receive  from  B  ? 


FRACTIONAL  EQUATIONS.  171 

4.  The  fore  wheel  of  a  carriage  is  a  feet  in  circumfer- 
ence, and  the  hind  wheel  is  b  feet.  Find  the  distance 
traveled  when  the  fore  wheel  has  made  c  revolutions  more 
than  the  hind  wheel. 

5.  Two  towns,  P  and  Q,  are  a  miles  apart.  One  person 
sets  out  from  P  and  travels  towards  Q  at  the  rate  of  b  miles 
an  hour,  and  at  the  same  time  another  person  sets  out  from 
Q  and  travels  towards  P  at  the  rate  of  c  miles  an  hour. 
How  many  miles  from  P  will  they  meet  ? 

6.  A  person  was  employed  a  days  on  these  conditions : 
for  each  day  he  worked  he  was  to  receive  b  cents,  and  for 
each  day  he  was  idle  he  was  to  forfeit  c  cents.  At  the  end 
of  a  days  he  received  d  cents.    How  many  days  did  he  work  ? 

7.  A  banker  has  two  kinds  of  coins :  it  takes  a  pieces  of 
the  first  to  make  a  dollar,  and  b  pieces  of  the  second  to  make 
a  dollar.  A  person  wishes  to  obtain  c  pieces  for  a  dollar. 
How  many  pieces  of  each  kind  must  the  banker  give  him  ? 

Interest  Formulas. 

178.  The  elements  involved  in  computation  of  interest 
are  the  principal,  rate,  time,  interest,  and  amount. 

Let  p  =  the  principal, 

r  =  the  interest  of  $1  for  1  year,  at  the  given  rate, 
y^f  =  the  time  expressed  in  years, 

i  =  the  interest  for  the  given  time  and  rate, 
Sa  =  the  amount  (sum  of  principal  and  interest). 

179.  Given  the  Principal,  Rate,  and  Time ;  to  Find  the  Interest. 

Since  r  is  the  interest  of  $  1  for  1  year,  pr  is  the  interest 
of  $p  for  1  year,  and  prt  is  the  interest  of  %p  for  t  years. 

.\  i  =  prt.  (Formula  1) 

Rule.     Find  the  product  of  the  principal,  rate,  and  time. 


172  FRACTIONAL  EQUATIONS. 

180.  Given  the  Interest,  Rate,  and  Time ;  to  Find  the  Principal, 

By  formula  1,  prt  =  i. 

Divide  by  rt,  p  =  —  •  (Formula  2) 

181.  Given  the  Amount,  Rate,  and  Time ;  to  Find  the  Principal. 

From  formula  1,    p  -\-prt  =  a, 
or  p  (1  -f-  rt)  =  a. 

a 

Divide  by  1  +  rt,  p  =  •  (Formula  3) 

182.  Given  the  Amount,  Principal,  and  Rate ;  to  Find  the  Time. 

From  formula  1,    p  +prt  =  a. 
Transpose  p}  prt  =  a  —p. 

Divide  by  pr,  t  =  — —'  (Formula  4) 

183.  Given  the  Amount,  Principal,  and  Time ;  to  Find  the  Rate. 

From  formula  1,    p  -\-prt  =  a. 
Transpose  p,  prt  =  a  —p. 

Divide  by  pt,  r  =  — £*•  (Formula  5) 

Exercise  69. 
Solve  by  the  preceding  formulas : 

1.  The  sum  of  two  numbers  is  40,  and  their  difference  is 
10.     Find  the  numbers. 

2.  The  sum  of  two  angles  is  100°,  and  their  difference  is 
21°  30'.     Find  the  angles. 

3.  The   sum  of   two  angles  is  116°  24'  30",  and  their 
difference  is  56°  2V  44".     Find  the  angles. 


FRACTIONAL  EQUATIONS.  173 

4.  A  can  do  a  piece  of  work  in  6  days,  and  B  in  5  days. 
How  long  will  it  take  both  together  to  do  it  ? 

5.  Find  the  interest  of  $2750  for  3  years  at  4£  per 
cent. 

6.  Find  the  interest  of  $950  for  2  years  6  months  at  5 
per  cent. 

7.  Find  the  amount  of  $2000  for  7  years  4  months  at  6 
per  cent. 

8.  Find  the  rate  if  the  interest  on  $680  for  7  months  is 
$35.70. 

9.  Find  the  rate  if  the  amount  of  $750  for  4  years  is 
$900. 

10.  Find  the  rate  if  a  sum  of  money  doubles  in  16  years 
8  months. 

11.  Find  the  time  required  for  the  interest  on  $2130  to 
be  $436.65  at  6  per  cent. 

12.  Find  the  time  required  for  the  interest  at  5  per  cent 
on  a  sum  of  money  to  be  equal  to  the  principal. 

13.  Find  the  principal  that  will  produce  $161.25  interest 
in  3  years  9  months  at  8  per  cent. 

14.  Find  the  principal  that  will  amount  to  $1500  in  3 
years  4  months  at  6  per  cent. 

15.  How  much  money  is  required  to  yield  $2000  interest 
annually  if  the  money  is  invested  at  5  per  cent  ? 

16.  Find  the  time  in  which  $640  will  amount  to  $1000 
at  6  per  cent. 

17.  Find  the  principal  that  will  produce  $100  per  month, 
at  6  per  cent. 

18.  Find  the  rate  if  the  interest  on  $700  for  10  months 
is  $25. 


CHAPTER  XL 
SIMULTANEOUS   SIMPLE  EQUATIONS. 

184.  If  we  have  two  unknown  numbers  and  but  one  rela- 
tion between  them,  we  can  find  an  unlimited  number  of 
pairs  of  values  for  which  the  given  relation  will  hold  true. 

Thus,  if  x  and  y  are  unknown,  and  we  have  given  only  the  one 
relation  x  +  y  =  10,  we  can  assume  any  value  for  x,  and  then  from 
the  relation  x  +  y  =  10  find  the  corresponding  value  of  y.  For  from 
x  +  y  =  10  we  find  y  =  10  —  x.  If  x  stands  for  1,  y  stands  for  9 ;  if 
x  stands  for  2,  y  stands  for  8 ;  if  x  stands  for  —  2,  y  stands  for  12 ; 
and  so  on  without  end. 

185.  We  may,  however,  have  two  equations  that  express 
different  relations  between  the  two  unknowns.  •  Such  equa- 
tions are   called   independent  equations. 

Thus,  x  +  y  =  10  and  x  —  y  =  2  are  independent  equations,  for 
they  evidently  express  different  relations  between  x  and  y. 

186.  Independent  equations  involving  the  sameunknowns 
are  called  simultaneous  equations. 

If  we  have  two  unknowns,  and  have  given  two  independ- 
ent equations  involving  them,  there  is  but  one  pair  of  values 
which  will  hold  true  for  both  equations. 

Thus,  if  in  §  184,  besides  the  relation  x  +  y  =  10,  we  have  also  the 
relation  x  —  y  =  2,  the  only  pair  of  values  for  which  both  equations 
will  hold  true  is  the  pair  x  =  6,  y  =  4. 

Observe  that  in  this  problem  x  stands  for  the  same  number  in  both 
equations ;  so  also  does  y. 


SIMULTANEOUS  SIMPLE  EQUATIONS.  175 

187.  Simultaneous  equations  are  solved  by  combining 
the  equations  so  as  to  obtain  a  single  equation  with  one 
unknown  number;  this  process  is  called  elimination. 

There  are  three  methods  of  elimination  in  general  use  : 

I.    By  Addition  or  Subtraction. 
II.    By  Substitution. 
III.    By  Comparison. 

188.  Elimination  by  Addition  or  Subtraction. 

1.  Solve:  5x-3y  =  20\  (1) 
2x  +  5y  =  39)  (2) 

Multiply  (1)  by  5,  and  (2)  by  3, 

25* -15?/ =  100  (3) 

6*  +  15y  =  117  (4) 

Add  (3)  and  (4),  31*  =217 

.-.  *  =  7. 

Substitute  the  value  of  *  in  (2), 

14  +  5y  =  39. 

•••  y  =  5. 

In  this  solution  y  is  eliminated  by  addition. 

2.  Solve:  6x  +  35y  =  177\  (1) 
Sx-21y  =    33)                            (2) 

Multiply  (1)  by  4,  and  (2)  by  3, 

24x  +  140y  =  708  (3) 

24*-    63 y=    99  (4) 

Subtract,  203  y  =  609 

.-.  y  =  3. 

Substitute  the  value  of  y  in  (2), 

8*  — 63  =  33. 

.-.  *  =  12. 

In  this  solution  x  is  eliminated  by  subtraction. 


176  SIMULTANEOUS  SIMPLE  EQUATIONS, 

189.   To  Eliminate  by  Addition  or  Subtraction,  therefore, 

Multiply  the  equations  by  such  numbers  as  will  make  the 
coefficients  of  one  of  the  unknown  numbers  equal  in  the 
resulting  equations. 

Add  the  resulting  equations,  or  subtract  one  from  the  other, 
according  as  these  equal  coefficients  have  unlike  or  like  signs. 

Note.  It  is  generally  best  to  select  the  letter  to  be  eliminated 
that  requires  the  smallest  multipliers  to  make  its  coefficients  equal ; 
and  the  smallest  multiplier  for  each  equation  is  found  by  dividing 
the  L.  C.  M.  of  the  coefficients  of  this  letter  by  the  given  coefficient  in 
that  equation.  Thus,  in  example  2,  the  L.  C.  M.  of  6  and  8  (the  co- 
efficients of  x)  is  24,  and  hence  the  smallest  multipliers  of  the  two 
equations  are  4  and  3,  respectively. 

Sometimes  the  solution  is  simplified  "by  first  adding  the 
given  equations  or  by  subtracting  one  from  the  other. 

Solve:  '       x  +  49  y  =  51)  (1) 

49a;  +      y  =  99  )  (2) 

Add  (1)  and  (2),  50  X  +  50  y  =  150.  (3) 

Divide  (3)  by  50,  X  +  y  =  3.  (4) 

Subtract  (4)  from  (1),  48  y  =  48. 

.-.  V  =  1. 
Subtract  (4)  from  (2),  48  x  =  96. 

.-.  x  =  2. 

Exercise  70. 
Solve  by  addition  or  subtraction: 

1.  5x  +  2y  =  39\  4.   4a-5y  =  26) 
2X-     y=    3>  Sx-6y  =  15i 

2.  x  +  3y  =  22\  5.      x  +  2y  =  35\ 
2x-4y=    4)  3x-2y  =  17$ 

3.  7x-2y  =  ll\  6.      x  +  4y  =  35\ 

x  +  5y  =  28  )  2x  -  3y  =  26  ) 


SIMULTANEOUS  SIMPLE  EQUATIONS.  177 


7. 

3x  +  5y  =  50\ 

11. 

«  +    2y  = 

91 

90) 

3x-    3y  = 

8. 

5x  +  2y  =  36\ 
2x  +  3y  =  43) 

12. 

4cc  —    3y  = 

39 1 
17) 

3x  —    4y  = 

9. 

3x  +  7y  =  50\ 
5x-2y  =  15)> 

13. 

7x-    2y  = 

69  "> 

39) 

x  —  10  y  = 

10. 

2x+    y  =    3\ 
lx  +  5y  =  2l) 

14. 

3a+    7y  = 

16  \ 
13) 

2z  +    5y  = 

190. 

Elimination  by  Substitution. 

Solv 

e:                     5sc  +  4 

y  =  32 

} 

(i) 

4z  +  3 

y  =  25 

(2) 

Transpose  4  y  in  (1), 

6x- 

=  32-4y. 

(3) 

Divide  by  coefficient  of 

X,           X: 

_32-4y_ 
5 

<*) 

Substitute  the  value  of : 

c  in  (2), 

- 

<(^)  +  *- 

=  25, 

128  - 16 
6 

V  +  3y-- 

=  25, 

128-165 

—  y- 

=  125, 
=  -3. 
=  3. 

Substitute  the  value  of  ? 

/in  (2), 

4»  +  9: 

=  25. 
=  4. 

To  Eliminate  by  Substitution,  therefore, 

From  one  of  the  equations  obtain  the  value  of  one  of  the 
cnknown  numbers  in  terms  of  the  other. 

Substitute  for  this  unknown  number  its  value  in  the  othei 
equation,  and  reduce  the  resulting  equation. 


178  SIMULTANEOUS   SIMPLE  EQUATIONS. 

Exercise  71. 

Solve  by  substitution : 

1.     2x-    ly=    0}  8.   3x-    2y  =  2S\ 

3x-    5y  =  ll)  2x  +    5y  =  63) 


Ax-    5y  =    4|  9.    2x-    3y  =  23~> 

Sx-    2y  =  10)  5x+    2y  =  2$i 

2x-    3y  =    1}  10.   6x-    7y  =  ll\ 

3a>-.2y  =  29)  5x-    6y  =    8) 


4.  x+      y  =  19\  11.   7x+    6y  =  20\ 

2x+    7y  =  88)  2x+    5y  =  32) 

5.  2x-       y=    5)  12.      «+    5y  =  37) 

x+    2y  =  25)  3x  +    2y  =  46) 

6.  19a;-15y  =  23">  13.   3cc-    7?/=  40) 
=  21)  4»-    3y=   9) 


13a;-    5^  =  21)  4a;-    3y 

jc  +  10y  =  73)  14.   5a>  +    9y 

7x-    2y=    7  S  3a;-f-lly 


191.   Elimination  by  Comparison. 

Solve:  2x-5y  =  66\  (1) 

3x  +  2y  =  23)  (2) 

Transpose  6  y  in  (1)  and  2  y  in  (2), 

2s  =  66  +  5y,  (8) 

8s  =  23-2y.  (4) 

Divide  (3)  by  2,  x  =  2S±*E  (5) 

Divide  (4)  by  8,  x  =  23~2?.  (6) 

t  *  66  +  5y      23  —  2y  _ 

Equate  the  values  of  x, = - — »  (7) 


SIMULTANEOUS  SIMPLE  EQUATIONS.  179 

Keduce  (7),  198  +  15  y  =  46  -  4  y, 

19  y  =  -  152. 

.-.y  =  -  8. 
Substitute  the  value  of  y  in  (1), 

2  x  +  40  =  06. 

.-.  x  =  13. 

192.    To  Eliminate  by  Comparison,  therefore, 

From  each  equation  obtain  the  value  of  one  of  the  unknown 
numbers  in  terms  of  the  other. 

Form  an  equation  from  these  equal  values  and  reduce  the 
equation. 

Exercise  72. 
Solve  by  comparison : 
1. 


2. 


4. 


6. 


7.    11 


X  +        y  =  30  \ 

Sx-    2y  =  25$ 

9. 

2a;-    3y=      lj 
5x+    2y  =  126i> 

?x+    Sy  =  70} 
5x  —    4=y  =    7  ) 

10. 

50a;-    9y  =      1} 
7x-       y=      3) 

9x  +    4?/  =  54| 
4a;  +    9y  =  S9i 

11. 

x  +  21y=      2 } 
27  ?/+    2  a;  =    19) 

7x  +    2y  =  6S} 
Sx-       y=    3) 

12. 

10a;  +    3?/  =  174| 
3a;  +  10?/  =  125) 

2  x  -33  ?/  =  29  > 

3  z  _  47  y  =  46  ) 

13. 

6a;-13y  =      2) 
5x-12y  =      4) 

2x-       y=    9} 
5X-    3yiUy 

14. 

2a;  +       2/  =  108]. 
10a;  +    2y=    60) 

la;—    7y=    6") 
9a;-    5y  =  10) 

15. 

3a?-    52/=      5\ 
7x+      y  =  2Q>5) 

8.      5a;  +  92/  =  188)  16.    12a;  +    7y=176| 

13a;-22/=    57)  3y-19a;=      3) 


180 


SIMULTANEOUS  SIMPLE  EQUATIONS. 


193.    Each  equation   must  be  simplified,  if  necessary, 
before  the  elimination. 


Solve:                      ia>-i(y  +  l)  =  l> 
i(s  +  l)  +  *(y-l)  =  9j 

(1) 

(2) 

Multiply  (1)  by  4,  and  (2)  by  12, 

Sx  —  2y  —  2  =  4, 

(3) 

4x  +  4  +  9y  —  9  =  108. 

(4) 

From  (3),                                   3x  —  2y  =  6. 

(5) 

From  (4),                                    4x  +  9y  =  113. 

(6) 

Multiply  (5)  by  4,  and  (6)  by  3, 

12x-    8y=    24 

(7) 

12x  +  27y  =  339 

(8) 

Subtract  (7)  from  (8),                       35  y  =  315 

.-.  y  =  9. 

Substitute  value  of  y  in  (1),                 x  =  8. 

Exercise  73. 


Solve : 


3^2      3 

-  +  £  =  ! 

2~r3      6 


*_1      £-2 

8^5 


2a;  -f 


2jr-5_ 


=  21 


2-        3      +      2 

2^      9 


4a;  +  5y 

=^+2„=i 


3z-5y      0      2s  +  y 
6-          2         Xo~      5 

0      a:-2y      a;  ,  y 
J               8           4      ~2+3       J 

6.    *£$**.*] 
x  —  y 

7* -13 

3^-5  "     J 

- 

7. 


SIMULTANEOUS  SIMPLE  EQUATIONS. 
+  1      y  +  2 ,2(a?-y) 


181 


3  4  5 

4  3  2 


g  +  2y  +  l 
°*    2x-y  +  l 

x-y  +  3 


x-2  10  -  x      y  -  10 
5  3      ~      4 

2y  +  4      4a  +  y  +  26 
3  8 


10. 


2a-y  +  3_a;-2y  +  19 

3  "  4 
3a:-4y  +  3_2y-4a;  +  21 

4  3 


11. 


7  +  sc 


^  =  3y-5 


12. 


5y  —  7   ,  4x  — 3       .Q      1 
2"  +  _ 6- 

26  +  5a:-6y 


13 


4y  —  3sc 


5a-6y      3(*  +  2y) 
1J+         6        ~  4 


to    *  +  8  3y-a        . 

13-  ^r  - 2      e 

2a;  +  y      9a?  — 7      7y-4a;  +  36 
2  8  16 


14. 


a;  +  2y  +  3  =  5y-4a;  — 6 
13  "  3 

6a?-5y  +  4  3a?  +  2y-H 
3  19 


182 


SIMULTANEOUS  SIMPLE  EQUATIONS. 


15. 


x  -f  y      5 
y  -x~  3 


16. 


y  —  4  _  #  4- 1 


„.  »£^l2  +  2«.3(j,-l) 

5a;  +  6y      4a;  —  3y  _ 

10  3  y 


18. 


4a;-3y-7      9a;-4y-25 


5 


30 


y-1      10a;-3y--20^3a;-f  2y  +  3 
3     +  20  ~  30 


Note.     In  solving  the  following  problems  proceed  as  in  §  175. 


6y  +  5      4a;-5y  +  3^9y-4 
8  4a; -2y        .      12 

8a;  +  3      a;-3y      6a;-l 
4       +   7-x  ~       3 


«**  2y  —  x 

20-*-23^  = 
-3 


y  + 


a -18 


a;-9£ 


4  a; -|- 7      5a;-4y  =  17  +  8s 

3       +  2a; +  8  ~~        6 
g  a;  -  12      4  a;  -  6  y  -  13  _  10  x  -  53 

4  2a;-3y      ~  8 


7  + 8a;      3(a;-2y)^ll4-4a; 

10  2  (a; -4)  5 

8(2y  +  3)^6y  +  21       3y-h5a; 
4  4  2(2y-3)J 


SIMULTANEOUS  SIMPLE  EQUATIONS.  183 

194.    Literal  Simultaneous  Equations. 

Solve  :  ax  +  by  =  c  > 

a'a;  +  6'?/  ==  c' ) 

Note.  The  letters  a',  b'  are  read  a  prime,  b  prime.  In  like  man- 
ner, a",  a"'  are  read  a  second,  a  third,  and  oi,  a2,  a3  are  read  a  sub 
one,  a  sub  two,  a  sub  three.  It  is  sometimes  convenient  to  represent 
different  numbers  that  have  a  common  property  by  the  same  letter 
marked  by  accents  or  suffixes.  Here  a  and  a'  have  a  common  prop- 
erty as  coefficients  of  x. 

ax  +by  =  c.  (1) 

a'x  +  b'y  =  c'.  (2) 

To  find  the  value  ofy,  multiply  (1)  by  a'  and  (2)  by  a. 
aafx  +  a'by  —  a'c 
aa'x  +  ab'y  =  ac' 


Subtract, 

Divide  by  a'b  — 

a'by  —  ab'y 
ab',                y 

—  a'c  - 
a'c  - 
a'b  - 

-  ac' 

-  ac' 
-ab' 

To  find  the  value 

of  x,  multiply  (1)  by 
ab'x  +  bb'y  —  b'c 
a'bx  -f  bb'y  =  be' 

b',  and 

(2)  by  b. 

Subtract, 

ab'x  —  a'bx 

=  b'c- 

-be' 

Divide  by  ab'  —  a'b, 


ab'  —  a'b 
Exercise  74. 


Solve 


1.  x  4-  y  =  s  >  5.    bx  -b  ay  =  abc ") 

x  —  y  =  d  )  x  =  dy  ) 

2.  race  +  ny  =  r  >  6.    foe  -f-  a?/  =  1  > 
ra'cc  -f-  wf?/  =  r'  )  b'x  —  a'y  =  1  ) 

3.  ax  —  by  =  c  ")  7.    Sbx  -\-  2  ay  =  3ab) 
a'x  +  #V  =  £  J  4:bx  —  3ay  =  %ab  ) 

4.  a>  —      y  =  raw ")  8.    2  cc  —  3y  =  a  —  b\ 
%cx  4-  aoy  =  ras)  3  cc  —  2y  =  a-r-b  ) 


184 


SIMULTANEOUS  SIMPLE  EQUATIONS, 
bx  cy  1 


9. 


a?-b2 


+ 


a  +  b 


10. 


bx  +  cy  =  a  +  b 

a  14.    ax  +  by  =  c 


y  —  n       b 
bx  +  ay  =  c 


a        o 
a       b~  3 


12. 


+ 


V 


a  +  b   '   a  —  b 
x  -\-  y  =  2a 


=  2 


x  —  y  =  a  —  b 


bx  +  ay 


:: 


15.    3a2  +    ax  =  b2  +  by 
ax  +  2by  =  d 


16. 


// 


17. 


a  -\-b      a  —  b      a  +  b  \ 

a-6j 
2/ 


a  +  b      a  —  b 


+  — 

m  —  a      m 


+ 


=  1 


1 


n  —  a      n 


195.  Fractional  simultaneous  equations,  of  which  the 
denominators  are  simple  expressions  and  contain  the  un- 
known numbers,  may  be  solved  as  follows: 


>olve :                       -  +  -  =  m 
x      y 

0) 

c    ,  d 

-  +  -  =  n 

x       y 

(2) 

To  find  the  value  of  y. 

Multiply  (1)  by  c,                1 =  cm. 

x       y 

(3) 

Multiply  (2)  by  a,               1 —  aw. 

x       y 

(*) 

Subtract  (4)  from  (3),        bc  ~  °^ 

=  cm  —  an. 

• 

SIMULTANEOUS  SIMPLE  EQUATIONS. 


185 


Multiply  both  sides  by  y,  and  we  have 

be  —  ad  =  (cm  —  an)  y. 

■~.  . ,    ■.  be  —  ad 

Divide  by  cm  —  em, 

To  find  the  value  ofx 
Multiply  (1)  by  a\ 

Multiply  (2)  by  6, 

Subtract  (6)  from  (5), 

Multiply  both  sides  by  x,  and  we  have 

ad  —  be  =  (dm  —  bn)  x. 
ad  — be 


y   _                       -    ■ 

cm  —  an 

ad  .  bd       , 

1 =  dm. 

x        y 

(5) 

be  ,    bd 

— 1 =  bn. 

x       y 

(6) 

ad  — be       .         , 
=  dm  —  bn. 

Divide  by  dm  —  bn, 
2.   Solve: 


dm  —  bn 


Sx       hy 

"7  1 

=  3 


(1) 

(2) 


6x      10  y 

Multiply  (1)  by  15,  the  L.  C.  M.  of  3  and  6 ;  and  (2)  by  30,  the 
C.  M.  of  6  and  10, 


^  +  «  =  105. 

x      y 

^-§  =  90. 

x       y 


(3) 
(4) 


Multiply  (4)  by  2,  and  add  the  result  to  (3), 


55  =  285. 

X 


Substitute  the  value  of  x  in  (1),  and  we  obtain 

1 


186 


SIMULTANEOUS  SIMPLE  EQUATIONS. 


Solve 


Exercise  75. 


1. 

M=2 

x       y 

x       y        ¥ 

2. 

«  +  ^  =  49 

x        y 

1  +  ^  =  23 
x      y 

3. 

2  +  «  =  3   1 

15_4  = 


1 

a;      y 

*-2  =  7 

a      y           j 

- 

5. 

A+ifh5" 

_5_     _4_  _.  11 

4x      5y      20 

6. 

2^+3i;=3 

4#      oy 

3.9 

7.   — I —  =  m 
x      y 


8.   ^  +  ±  =  b 
x      y 

n      1 
-4-  -  =  c 


m 

■  w 

9. 

+  -  = 

=  a 

x 

y 

r 

.     5 

-h-  = 

b 

X 

2/ 

a 

6 

ac 

0. 

— 

—  —  — 

:  

X 

y 

b 

b 

a 

be 

X 

y~ 

a 

11. 


ax 


2^ 

by 


ax      by 


12.  ±  +  JL  =  a  +  h 

bx      ay 


b.a         2 
-  +  -  =  a2 

*     y 


SIMULTANEOUS  SIMPLE  EQUATIONS. 


187 


196.  If  three  simultaneous  equations  are  given,  involving 
three  unknown  numbers,  one  of  the  unknowns  must  be 
eliminated  between  two  pairs  of  the  equations;  then  a 
second  unknown  between  the  two  resulting  equations. 

Likewise,  if  four  or  more  equations  are  given,  involving 
four  or  more  unknown  numbers,  one  of  the  unknowns  must 
be  eliminated  between  three  or  more  pairs  of  the  equations ; 
then  a  second  between  the  pairs  that  can  be  formed  of  the 
resulting  equations ;   and  so  on. 

Note.  The  pairs  chosen  to  eliminate  from  must  be  independent 
pairs,  so  that  each  of  the  given  equations  shall  be  used  in  the  process 
of  the  eliminations. 


Solve:  2x  -3y  +  4«=    4" 

5x  —    y  —  &z  =    5. 

Eliminate  z  between  the  equations  (1)  and  (3). 

Multiply  (1)  by     2,  4x  -  Qy  +  82  =    8 

(3)  is  5x  —    y  —  %z=   5 

Add,  9x-7y  =13 

Eliminate  z  between  the  equations  (1)  and  (2). 

Multiply  (1)  by  7,         14  x  -  21  y  +  28  z  =  28 


Multiply  (2)  by  4, 
Add, 


12s  +  20y-28z  =  48 


26  x 


y 


=  76 


(1) 
(2) 
(3) 

(4) 
(5) 


(6) 


We  now  have  two  equations  (5)  and  (6)  involving  two  un- 
knowns, x  and  y. 

Multiply  (6)  by  7,  182  x  -  7  y  =  532  (7) 

(5)  is  9x-7y  =,  13 

Subtract  (5)  from  (7),  173  x 


Substitute  the  value  of  x  in  (6), 


=  519 
.-.  x  =  3. 
78  -  y  =  76. 

.-.  y  =  2. 


Substitute  the  values  of  x  and  y  in  (1), 

6-6  +  42  =  4. 
.%  *  m  % 


188  SIMULTANEOUS  SIMPLE  EQUATIONS. 


Solve: 

Exercise  76. 

1.     x  +  y-    8  =  0^1 

10.      5x  +  2y-20z  =  20 

y  +  *  -  28  =  0 

3x-6y  +    7z  =  51 

<c-fs-14  =  oJ 

4x  +  8y-    9*  =  53 

6. 


2.  4a  +  32/  +  2«  =  25^j  11.      x  +  2y  +  10z  =    44"i 
3a-2y  +  5s  =  20  J>  3a-f-3y  +    7z  =  384f- 

10x-5y  +  3*  =  17J  2a  +    y  +       s  =  256J 

3.  5o;-2y- 2^  =  12^1  12.      10a;  =     y  +  ±z  +  56^ 

#  +    y  +     *  =    8  f  3y  =  2z  +  3s-98  J> 

7x  +  3y  +  4z  =  42J  2*  =    ic-3y-18J 

4.  x-     y+     «=Hi  13.        3«-5y-2«  =  14^ 
3z  +  32/-2z  =  60  f-  5^-8?/-     z  =  12  I 

10x-5y-3«=    OJ  x-3y-3z  =    lJ 

5.  10  a-     2/  +  3«  =  42^  14.        2x  +  3y+     z  =  Sl^i 

7^  +  2*/  +     s  =  51f  a;-     y  +  3*  =  13l 

3a  +  3y-     z  =  24  J  lOy  +  5cc  -  2*  =  48J 


5x  +  2y-  3z  =  160^  15.      2cc  +    3y-    4z  =  l 
3a  +  9#  +  8s  =  115  >-  10x-    6y  +  12z  =  6 

2x-3y-5z=    45J  x  +  12y+    2z  =  5. 

7.  6x-2y  +  5z  =  53^  16.      3a  +    6y  +    2z  =  3" 
5x  +  3y  +  7z  =  33>  12y+    4s-    6a  =  2 

cc  4"     2/ +     2  =    5J  9#-|-18y  —    4s  =  4. 

8.  3x-  3y'+  42  =  20^  17*  2x  +  V  +  2z==S 
6x  +  2y-7z  =  5>  5y-4:X-4:Z  =  l 
2x-     y  +  $z  =  4:5J  3x  +  9y+     z  =  9 

9.  2x  +  7y  +  10z  =  25^  18.    3x  +  2y+       z  =  20^ 

a+     y_       «=    9^  2x-     y+    3*  =  26£  V 

7s-7y-ll*  =  73J  a+    y  +  10*  =  55  J 


SIMULTANEOUS  SIMPLE  EQUATIONS. 


189 


19.    ---+    4  =  0 
x      y 

1-1  +    1  =  0 

y       z 

Z        X 


24.    1  +  1-1-0 

x      y      a 

i+l-i=o 

X         Z         0 

1+1-1=0 


.   1  +  i  +  i    =36 
x      y      z 

M-i    =28 

x      y      z 

l  +  f  +  f  =  20 

x     3y     2z 


2,   1  +  |  +  |=62 


.!+?-«=  i 

x      y      z 

5  +  1  +  ^  =  24 

x      y      z 


26.    5  +  1-2  =  0 

a;      y      * 

2-5-2=0 

*        2/ 

1+1-^=0 
as      z      3 


«.  1+1-1=. 

a      y      z 

x      y      z 

1^1       1 
-H =  c 

y      z      x 


„.   «_4  +  5=    38 

x       y      z 

5  +  2  +  15=    61 

x      y       z 

*-«  +  *2  =  161 

a;      y       z 


53. 


4      3 

1 

x      y 

20 

2      3 

1 

—  —  —  ^^ 

— 

£         £C 

15 

4      5_ 

1 

«      2/ 

12 

j 

28.  5_§  +  4  =  29 

a:      y      * 

5_6_r  =  _104 

a;      y      * 

9  +  10_8  =  U9 

y        z       x 


CHAPTER  XH. 

PROBLEMS  INVOLVING  TWO  OR  MORE  UNBLOWN 
NUMBERS. 

197.  It  is  often  necessary  in  the  solution  of  problems  to 
employ  two  or  more  letters  to  represent  the  numbers  to  be 
found.  In  all  cases  the  conditions  must  be  sufficient  to 
give  just  as  many  equations  as  there  are  unknown  numbers 
employed.  If  there  are  more  equations  than  unknown 
numbers,  some  of  them  are  superfluous  or  inconsistent ;  if 
there  are  fewer  equations  than  unknown  numbers,  the 
problem  is  indeterminate. 

Exercise  77. 

1.  If  A  gave  B  $10,  B  would  have  three  times  as  much 
money  as  A.  If  B  gave  A  $10,  A  would  have  twice  as  much 
money  as  B.     How  much  has  each  ? 

Let  x  =  the  number  of  dollars  A  has, 

and  y  =  the  number  of  dollars  B  has. 

Then,  if  A  gave  B  $10, 

x  —  10  =  the  number  of  dollars  A  would  have, 
y  +  10  =  the  number  of  dollars  B  would  have. 
.-.  y  +  10  =  3(x-10).  (1) 

If  B  gave  A  $10, 

x  +  10  =  the  number  of  dollars  A  would  have, 
y  —  10  =  the  number  of  dollars  B  would  have. 
.-.x  +  10  =  2(y-10).  (2) 

From  the  solution  of  equations  (1)  and  (2),  *  =  22,  and  y  =  26. 

Therefore,  A  has  $22,  and  B  has  $26. 


PROBLEMS.  191 

2.  If  the  smaller  of  two  numbers  is  divided  by  the 
greater,  the  quotient  is  0.21,  and  the  remainder  0.0057; 
but  if  the  greater  is  divided  by  the  smaller,  the  quotient  is 
4  and  the  remainder  0.742.     Find  the  numbers. 

Let  x  =  the  greater  number, 

and  y  =  the  smaller  number. 

Then,  Inmost  =021)  (1) 

X 

x  —  0.742 
and  =  4.  (2) 

.-.  y  -  0.21  x  =  0.0067,  (3) 

x-4y  =  0.742.  (4) 

Multiply  (3)  by  4,  4  y  —  0.84  x  =  0.0228  (6) 

(4)  is  —  4y+         x  =  0.742 

Add,  0.16  x  =  0.7648 

.-.  x  =  4.78. 
Put  the  value  of  x  in  (2),  4  y  =  4.038, 

.-.  y  =  1.0095. 

Therefore,  the  numbers  are  4.78  and  1.0095. 

3.  If  A  gave  B  $  100,  A  would  then  have  half  as  much 
money  as  B ;  but  if  B  gave  A  $100,  B  would  have  one  third 
as  much  as  A.     How  much  has  each  ? 

4.  If  the  greater  of  two  numbers  is  divided  by  the 
smaller,  the  quotient  is  7  and  the  remainder  4;  but  if  three 
times  the  greater  number  is  divided  by  twice  the  smaller, 
the  quotient  is  11  and  the  remainder  4.     Find  the  numbers. 

5.  If  the  greater  of  two  numbers  is  divided  by  the 
smaller,  the  quotient  is  4  and  the  remainder  0.37;  but  if 
the  smaller  is  divided  by  the  greater,  the  quotient  is  0.23 
and  the  remainder  0.0149.     Find  the  numbers. 

6.  If  A  gave  B  $5,  he  would  have  $  6  less  than  B ;  but  if 
he  received  $5  from  B,  three  times  his  money  would  be  $20 
more  than  four  times  B's.     How  much  has  each  ? 


192  PROBLEMS. 

7.  If  the  numerator  of  a  fraction  is  doubled  and  its 
denominator  diminished  by  1,  its  value  will  be  \.  If  its 
denominator  is  doubled  and  its  numerator  increased  by  1, 
its  value  will  be  \.     Find  the  fraction. 

Let  x  =  the  numerator, 

and  y  =  the  denominator. 

Then,  j-^L  =  ,,  (1) 

and  2_L1  =  |.  (2) 

The  solution  of  equations  (1)  and  (2)  gives  5  for  x  and  21  for  y. 
Therefore,  the  required  fraction  is  /T. 

8.  A  certain  fraction  becomes  equal  to  \  if  3  is  added 
to  its  numerator  and  1  to  its  denominator,  and  equal  to  \ 
if  3  is  subtracted  from  its  numerator  and  from  its  denomi- 
nator.    Find  the  fraction. 

9.  A  certain  fraction  becomes  equal  to  T9T  if  1  is  added 
to  double  its  numerator,  and  equal  to  £  if  3  is  subtracted 
from  its  numerator  and  from  its  denominator.  Find  the 
fraction. 

10.  There  are  two  fractions  with  numerators  11  and  5, 
respectively,  whose  sum  is  If ;  but  if  their  denominators 
are  interchanged  their  sum  is  2J.     Find  the  fractions. 

11.  A  certain  fraction  becomes  equal  to  \  when  7  is  added 
to  its  denominator,  and  equal  to  2  when  13  is  added  to  its 
numerator.     Find  the  fraction. 

12.  A  certain  fraction  becomes  equal  to  £  when  the 
denominator  is  increased  by  4,  and  equal  to  f  £  when  the 
numerator  is  diminished  by  15.     Find  the  fraction. 

13.  A  certain  fraction  becomes  equal  to  §  if  7  is  added 
to  the  numerator,  and  equal  to  §  if  7  is  subtracted  from 
the  denominator.     Find  the  fraction. 


PROBLEMS.  193 

14.  A  certain  number  is  expressed  by  three  digits.  The 
sum  of  the  digits  is  21.  The  sum  of  the  first  and  last  digits 
is  twice  the  middle  digit.  If  the  hundreds'  and  tens'  digits 
are  interchanged,  the  number  is  diminished  by  90.  Find 
the  number. 

Let  x  =  the  hundreds'  digit, 

y  =  the  tens'  digit, 
z  =  the  units'  digit. 
Then,        100  x  +  10  y  +  z  =  the  number. 

By  the  conditions,  x  +  y  +  z  =  21,  (1) 

x  +  z  =  2y,  (2) 

and  1002/  +  10z  +  z  =  100x+  10y  +  z-90.      (3) 

Solving  these  equations,  x  =  8,  y  =  7,  z  =  6. 

Therefore,  the  number  is  876. 

15.  The  sum  of  the  two  digits  of  a  number  is  9,  and  if 
27  is  subtracted  from  the  number,  the  digits  will  be  reversed. 
Find  the  number. 

16.  The  sum  of  the  two  digits  of  a  number  is  9,  and  if 
the  number  is  divided  by  the  sum  of  the  digits,  the  quotient 
is  5.     Find  the  number. 

17.  A  certain  number  is  expressed  by  two  digits.  The 
sum  of  the  digits  is  11.  If  the  digits  are  reversed,  the  new 
number  exceeds  the  given  number  by  27.     Find  the  number. 

18.  A  certain  number  is  expressed  by  three  digits,  the 
units'  digit  being  zero.  If  the  hundreds'  and  tens'  digits 
are  interchanged,  the  number  is  diminished  by  180.  If  the 
hundreds'  digit  is  halved,  and  the  tens'  and  units'  digits  are 
interchanged,  the  number  is  diminished  by  336.  Find  the 
number. 

19.  A  number  is  expressed  by  three  digits.  If  the  digits 
are  reversed,  the  new  number  exceeds  the  given  number  by 
99.  If  the  number  is  divided  by  nine  times  the  sum  of  its 
digits,  the  quotient  is  3.  The  sum  of  the  hundreds'  and 
units'  digits  exceeds  the  tens'  digit  by  1.     Find  the  number 


194  PROBLEMS. 

20.  A  boatman  rows  20  miles  down  a  river  and  back  in 
8  hours.  He  finds  that  he  can  row  5  miles  down  the  river 
in  the  same  time  that  he  rows  3  miles  up  the  river.  Find 
the  time  he  was  rowing  down  and  up  respectively. 

Let  x  =  the  boatman's  rate  per  hour  in  still  water, 

and  y  =  the  rate  per  hour  of  the  current. 

20 
Then,  ■  =  the  number  of  hours  he  was  rowing  down, 

20 
and  _     =  the  number  of  hours  he  was  rowing  up. 

Therefore,  -~-  +  -^-  =  8,  (1) 

x  +  y      x—y  v ' 

5  3 

and  — I —  = (2) 

x  +  y      x  —  y  v  ' 

Solving  these  equations,  x  =  5i,  y  =  1£. 

20  20 

Therefore,  — ; —  =  3,  =  5. 

x+y  x—y 

It  takes  him  3  hours  to  row  down  and  5  hours  to  row 
back. 

21.  A  boat's  crew  which  can  pull  down  a  river  at  the 
rate  of  10  miles  an  hour  finds  that  it  takes  twice  as  long  to 
row  a  mile  up  the  river  as  to  row  a  mile  down.  Find  the 
rate  of  their  rowing  in  still  water  and  the  rate  of  the 
stream. 

22.  A  boatman  rows  down  a  stream,  which  runs  at  the 
rate  of  2\  miles  an  hour,  for  a  certain  distance  in  1  hour 
30  minutes;  it  takes  him  4  hours  30  minutes  to  return. 
Find  the  distance  he  pulled  down  the  stream  and  his  rate 
of  rowing  in  still  water. 

23.  A  person  rows  down  a  stream  a  distance  of  20  miles 
and  back  again  in  10  hours.  He  finds  he  can  row  2  miles 
against  the  stream  in  the  same  time  he  can  row  3  miles  with 
it.  Find  the  time  of  his  rowing  down  and  of  his  rowing 
up  the  stream ;  and  also  the  rate  of  the  stream. 


PROBLEMS.  195 

24.  A  and  B  can  do  a  piece  of  work  together  in  3  days, 
A  and  C  in  4  days,  B  and  C  in  4£  days.  How  long  will  it 
take  each  alone  to  do  the  work  ? 

Let  x,  y,  z  =  the  number  of  days  in  which  A,  B,  C  can  do  the 
work,  respectively. 

Then,  -*-»-  =  the  parts  A,  B,  C  can  do  in  1  day,  respectively. 

And       -  +  -  =  the  part  A  and  B  together  can  do  in  one  day. 
x      y 

But  £  =  the  part  A  and  B  together  can  do  in  1  day. 

Therefore,  1  +  1  =  1,  (1) 

x      y      3  v  ' 

Likewise,  -  +  -  =  -,  (2) 

x      z      4 

and  1  +  1=4    (li  =  l)  <3> 

y      z      9      \4£      9/  *  ' 

Add,  and  divide  by  2,  -  +  -  +  -  =  f£  (4) 

x      y      z      7 2 

Subtract  (1),  (2),  and  (3),  separately  from  (4),  and  we  have 

!=-JL  i_ii  !-.!§. 

z      72'  y      72'  x      72* 
Therefore,  z  =  14f ,  y  -  6T6T,  a;  =  5T\. 

Therefore,  A  can  do  the  work  in  5 fa  days,  B  in  6/j 
days,  and  C  in  14f  days. 

25.  A  cistern  has  three  pipes,  A,  B,  and  C.  A  and  B 
will  fill  the  cistern  in  1  hour  10  minntes,  A  and  C  in  1 
hour  24  minutes,  B  and  C  in  2  hours  20  minutes.  How  long 
will  it  take  each  pipe  alone  to  fill  it  ? 

26.  A  and  B  can  do  a  piece  of  work  in  2\  days,  A  and 
C  in  3^  days,  B  and  C  in  4  days.  How  long  will  it  take 
each  alone  to  do  the  work  ? 

27.  A  and  B  can  do  a  piece  of  work  in  a  days,  A  and  C 
in  b  days,  B  and  C  in  c  days.  How  long  will  it  take  each 
alone  to  do  the  work  ? 


196  PROBLEMS. 

28.  A  sum  of  money,  at  simple  interest,  amounted  to 
$2480  in  4  years,  and  to  $2600  in  5  years.  Find  the  sum 
and  the  rate  of  interest. 

Let  x  =  the  number  of  dollars  in  the  principal, 

and  y  =  the  rate  of  interest. 

The  interest  for  one  year  is  r^r  of  the  principal,  =  -r^r  of  x ;  for  4 
100  100 

years,  =  —  of  x ;  and  for  5  years,  =  — ^  of  x. 
The  amount  is  principal  +  interest. 
Therefore,  x  +  |g|  =  2480. 

*  +  |H  =  2aoo. 

Hence,  100*  +  4«y  =  248,000.  (1) 

100  x  +  5  xy  =  260,000.  (2) 

Multiply  (1)  by  5  and  (2)  by  4,  and  we  have 

500x  +  20xy  =  1,240,000 

400x  +  20xy  =  1,040,000 
Subtract,  100  x  =     200,000. 

Therefore,  x  =  2000. 

Substitute  the  value  of  x  in  (1),  y  =  6. 
Therefore,  the  sum  is  $2000,  and  the  rate  6%. 

29.  A  sum  of  money,  at  simple  interest,  amounted  in  4 
years  to  $29,000,  and  in  5  years  to  $30,000.  Find  the  sum 
and  the  rate  of  interest. 

30.  A  sum  of  money,  at  simple  interest,  amounted  in  10 
months  to  $2100,  and  in  18  months  to  $2180.  Find  the 
sum  and  the  rate  of  interest. 

31.  A  man  has  $10,000  invested.  For  a  part  of  this 
sum  he  receives  5  per  cent  interest,  and  for  the  rest  6  per 
cent;  the  income  from  his  5  per  cent  investment  is  $60 
more  than  from  his  6  per  cent.  How  much  has  he  in  each 
investment  ? 


PROBLEMS.  197 

32.  In  a  mile  race  A  gives  B  a  start  of  20  yards  and 
beats  him  by  30  seconds.  At  the  second  trial  A  gives  B  a 
start  of  32  seconds  and  beats  him  by  9^T  yards.  Find  the 
number  of  yards  each  runs  a  second. 

Let  x  =  the  number  of  yards  A  runs  a  second, 

and  y  =  the  number  of  yards  B  runs  a  second. 

Since  there  are  1760  yards  in  a  mile, 

=  the  number  of  seconds  it  takes  A  to  run  a  mile. 

x 

Since  B  has  a  start  of  20  yards,  he  runs  1740  yards  the  first  trial  ; 

and  as  he  was  30  seconds  longer  than  A, 

h  30  =  the  number  of  seconds  B  was  running. 

x 

But        =  the  number  of  seconds  B  was  running. 

,!™  =  «29  +  80.  (1) 

y  x  v  ' 

In  the  second  trial  B  runs  (1760  —  9T5T)  yards  =  1750T6T  yards. 

,.l™&=l™  +  32.  (2) 

y  x  w 

From  the  solution  of  equations  (1)  and  (2),  x  =  5}f ,  and  y  =  5T3T. 

Therefore,  A  runs  5|f  yards  a  second,  and  B  runs  5^ 
rards  a  second. 

33.  Two  men,  A  and  B,  run  a  mile,  and  A  wins  by  2 
seconds.  In  the  second  trial  B  has  a  start  of  18J  yards, 
and  wins  by  1  second.  Find  the  number  of  yards  each 
runs  a  second,  and  the  number  of  miles  each  would  run  in 
an  hour. 

34.  In  a  mile  race  A  gives  B  a  start  of  3  seconds,  and  is 
beaten  by  12^  yards.  In  the  second  trial  A  gives  B  a  start 
of  10  yards,  and  the  race  is  a  tie.  Find  the  number  of 
yards  each  runs  a  second.  At  this  rate,  how  many  miles 
could  each  run  in  an  hour  ? 


198  PROBLEMS. 

35.  A  train,  after  traveling  an  hour  from  A  towards  B, 
meets  with  an  accident  which  detains  it  half  an  hour ;  after 
which  it  proceeds  at  four  fifths  of  its  usual  rate,  and  arrives 
an  hour  and  a  quarter  late.  If  the  accident  had  happened 
30  miles  farther  on,  the  train  would  have  been  only  an  hour 
late.     Find  the  usual  rate  of  the  train. 

Let  y  =  the  number  of  miles  from  A  to  B, 

and  5  x  =  the  number  of  miles  the  train  travels  per  hour. 

Then,  4  x  =  the  rate  of  the  train  after  the  accident. 

Then,     y  —  5  x  =  the  number  of  miles  the  train  has  to  go  after 
the  accident. 

Hence,    *—z —  the  number  of  hours  required  usually, 

5  x 

and  *— =  the  number  of  hours  actually  required. 

y  —  bx      y  —  5x  ' ' .       .,  , 

.*.  — 2— =  the  loss  in  hours  of  running  time. 

4a;  bx 

But  since  the  train  was  detained  £  an  hour  and  arrived  1£  hours 
late,  the  running  time  was  £  of  an  hour  more  than  usual.     That  is, 
£  =  loss  in  hours  of  running  time. 

y  —  bx      y  —  bx_S  m 

*'•     4x  bx         4'  K} 

If  the  accident  had  happened  30  miles  farther  on,  the  remainder  of 
the  journey  would  have  been  y  —  (5  x  +  30)  miles,  and  the  loss  in 
running  time  would  have  been  f  an  hour. 

y  -  (5s  +  30)      y  -  (5x  +  30)  _  1 , 

4x  bx  2  K  ' 

From  the  solution  of  equations  (1)  and  (2),  x  =  6,  and  bx  =  30. 

Therefore,  the  usual  rate  of  the  train  is  30  miles  an  hour. 

36.  An  express  train,  after  traveling  an  hour  from  A 
towards  B,  meets  with  an  accident  which  delays  it  15  min- 
utes. It  afterwards  proceeds  at  two  thirds  its  usual  rate, 
and  arrives  24  minutes  late.  If  the  accident  had  happened 
5  miles  farther  on,  the  train  would  have  been  only  21 
minutes  late.     Find  the  usual  rate  of  the  train. 


PROBLEMS.  199 

37.  If  3  yards  of  velvet  and  12  yards  of  silk  cost  $60, 
and  4  yards  of  velvet  and  5  yards  of  silk  cost  $58,  what  is 
the  price  of  a  yard  of  velvet  and  of  a  yard  of  silk  ? 

&8.  If  5  bushels  of  wheat,  4  of  rye,  and  3  of  oats  are  sold 
for  $9 ;  3  bushels  of  wheat,  5  of  rye,  and  6  of  oats  for  $8.75 ; 
and  2  bushels  of  wheat,  3  of  rye,  and  9  of  oats  for  $7.25 ; 
what  is  the  price  per  bushel  of  each  kind  of  grain  ? 

39.  A  train  proceeded  a  certain  distance  at  a  uniform 
rate.  If  the  speed  had  been  6  miles  an  hour  more,  the  time 
occupied  would  have  been  5  hours  less ;  but  if  the  speed  had 
been  6  miles  an  hour  less,  the  time  occupied  would  have 
been  1\  hours  more.     Find  the  distance. 

Hint.  If  x  =  the  number  of  hours  the  train  travels,  and  y  the 
number  of  miles  per  hour,  then  xy  =  the  distance. 

40.  A  certain  number  of  persons  paid  a  bill.  If  there  had 
been  10  persons  more,  each  would  have  paid  $2  less ;  but  if 
there  had  been  5  persons  less,  each  would  have  paid  $2.50 
more.   Find  the  number  of  persons  and  the  amount  of  the  bill. 

41.  A  man  bought  10  cows  and  50  sheep  for  $750.  He 
sold  the  cows  at  a  profit  of  10  per  cent,  and  the  sheep  at  a 
profit  of  30  per  cent,  and  received  in  all  $875.  Find  the 
average  cost  of  a  cow  and  of  a  sheep. 

42.  It  is  40  miles  from  Dover  to  Portland.  A  sets  out 
from  Dover,  and  B  from  Portland,  at  7  o'clock  a.m.,  to  meet 
each  other.  A  walks  at  the  rate  of  3^  miles  an  hour,  but 
stops  1  hour  on  the  way ;  B  walks  at  the  rate  of  2£  miles 
an  hour.  At  what  time  of  day  and  how  far  from  Portland 
will  they  meet  ? 

43.  A  number  is  expressed  by  three  digits.  The  sum  of 
the  digits  is  21 ;  the  sum  of  the  first  and  second  exceeds  the 
third  by  3 ;  and  if  198  is  added  to  the  number,  the  digits  in 
the  units'  and  hundreds'  places  will  be  interchanged. 
Find  the  number. 


200  PROBLEMS. 

44.  If  the  length  of  a  rectangular  field  is  increased  by  5 
yards  and  its  breadth  by  10  yards,  its  area  is  increased  by 
450  square  yards ;  but  if  its  length  is  increased  by  5  yards 
ana  its.  breadth  diminished  by  10  yards,  its  area  is  dimin- 
ished by  350  square  yards.     Find  its  dimensions. 

45.  If  the  floor  of  a  certain  hall  had  been  2  feet  longer 
and  4  feet  wider,  it  would  have  contained  528  square  feet 
more ;  but  if  the  length  and  width  were  each  2  feet  less,  it 
would  contain  316  square  feet  less.     Find  its  dimensions. 

46.  If  the  length  of  a  rectangle  was  4  feet  less  and  the 
width  3  feet  more,  the  figure  would  be  a  square  of  the  same 
area  as  the  given  rectangle.  Find  the  dimensions  of  the 
rectangle. 

47.  If  a  certain  number  is  divided  by  the  sum  of  its  two 
digits  diminished  by  2,  the  quotient  is  5  and  the  remainder 
1 ;  if  the  digits  are  interchanged,  and  the  resulting  number 
is  divided  by  the  sum  of  the  digits  increased  by  2,  the 
quotient  is  5  and  the  remainder  8.     Find  the  number. 

48.  A  person  has  a  certain  capital  invested  at  a  certain 
rate  per  cent.  Another  person  has  $2000  more  capital 
invested  at  one  per  cent  better  than  the  first,  and  receives 
$150  more  income.  A  third  person  has  $3000  more  capital 
invested  at  two  per  cent  better  than  the  first,  and  receives 
$280  more  income.  Find  the  capital  of  each,  and  the  rate 
at  which  it  is  invested. 

49.  A  man  makes  an  investment  at  4  per  cent,  and  a 
second  investment  at  4^  per  cent.  His  income  from  the 
two  investments  is  $715.  If  the  first  investment  had  been 
made  at  4£  per  cent  and  the  second  at  4  per  cent,  his  income 
would  have  been  $15  greater.  Find  the  amount  of  each 
investment 


PROBLEMS.  201 

50.  A  number  is  expressed  by  two  digits,  the  units'  digit 
being  the  larger.  If  the  number  is  divided  by  the  sum  of 
its  digits,  the  quotient  is  4.  If  the  digits  are  reversed  and 
the  resulting  number  is  divided  by  2  more  than  the  differ- 
ence of  the  digits,  the  quotient  is  14.     Find  the  number. 

51.  An  income  of  $335  a  year  is  obtained  from  two  in- 
vestments, one  in  4^  per  cent  stock  and  the  other  in  5  per 
cent  stock.  If  the  4£  per  cent  stock  should  be  sold  at  110, 
and  the  5  per  cent  at  125,  the  sum  realized  from  both  stocks 
together  would  be  $8300.    How  much  of  each  stock  is  there  ? 

52.  A  sum  of  money,  at  simple  interest,  amounted  in  m 
years  to  c  dollars,  and  in  n  years  to  d  dollars.  Find  the 
sum  and  the  rate  of  interest. 

53.  A  sum  of  money,  at  simple  interest,  amounted  in  m 
months  to  a  dollars,  and  in  n  months  to  b  dollars.  Find 
the  sum  and  the  rate  of  interest. 

54.  A  person  has  $18,375  to  invest.  He  can  buy  3  per 
cent  bonds  at  75,  and  5  per  cent  bonds  at  120.  How  much 
of  his  money  must  he  invest  in  each  kind  of  bonds  in  order 
to  have  the  same  income  from  each  investment  ? 

Hint.     Notice  that  the  3  per  cent  bonds  at  75  pay  4  per  cent  on  the 
loney  invested,  and  5  per  cent  bonds  at  120  pay  4|  per  cent. 

55.  In  a  mile  race  A  gives  B  a  start  of  44  yards,  and  is 
beaten  by  1  second.  In  a  second  trial  A  gives  B  a  start  of 
6  seconds,  and  beats  him  by  9£  yards.  Find  the  number 
of  yards  each  runs  a  second. 

56.  A  train,  after  running  2  hours  from  A  towards  B, 
meets  with  an  accident  which  delays  it  20  minutes.  It 
afterwards  proceeds  at  four  fifths  its  usual  rate,  and  arrives 
1  hour  40  minutes  late.  If  the  accident  had  happened 
40  miles  nearer  A,  the  train  would  have  been  2  hours  late. 
Find  the  usual  rate  of  the  train. 


202  PROBLEMS. 

57.  A  boy  bought  some  apples  at  3  for  5  cents,  and 
some  at  4  for  5  cents,  paying  $1  for  the  whole.  He  sold 
them  at  2  cents  apiece,  and  cleared  40  cents.  How  many 
of  each  kind  did  he  buy  ? 

58.  Find  the  area  of  a  rectangular  floor,  such  that  if  3 
feet  were  taken  from  the  length  and  3  feet  added  to  the 
breadth,  its  area  would  be  increased  by  6  square  feet,  but 
if  5  feet  were  taken  from  the  breadth  and  3  feet  added  to 
the  length,  its  area  would  be  diminished  by  90  square  feet. 

59.  A  courier  was  sent  from  A  to  B,  a  distance  of  147 
miles.  After  7  hours,  a  second  courier  was  sent  from  A, 
who  overtook  the  first  just  as  he  was  entering  B.  The  time 
required  by  the  first  to  travel  17  miles  added  to  the  time 
required  by  the  second  to  travel  76  miles  is  9  hours 
40  minutes.     How  many  miles  did  each  travel  per  hour  ? 

60.  A  box  contains  a  mixture  of  6  quarts  of  oats  and  9 
of  corn,  and  another  box  contains  a  mixture  of  6  quarts  of 
oats  and  2  of  corn.  How  many  quarts  must  be  taken  from 
each  box  in  order  to  have  a  mixture  of  7  quarts,  half  oats 
and  half  corn  ? 

61.  A  train  traveling  30  miles  an  hour  takes  21  minutes 
longer  to  go  from  A  to  B  than  a  train  which  travels  36 
miles  an  hour.     Find  the  distance  from  A  to  B. 

62.  A  man  buys  570  oranges,  some  at  16  for  25  cents, 
and  the  rest  at  18  for  25  cents.  He  sells  them  all  at  the 
rate  of  15  for  25  cents,  and  gains  75  cents.  How  many  of 
each  kind  does  he  buy  ? 

63.  A  and  B  run  a  mile  race.  In  the  first  heat  B 
receives  12  seconds  start,  and  is  beaten  by  44  yards.  In 
the  second  heat  B  receives  165  yards  start,  and  arrives  at 
the  winning  post  10  seconds  before  A.  Find  the  time  in 
which  each  can  run  a  mile. 


PROBLEMS.  203 

198.  The  discussion  of  a  problem  consists  in  making 
various  suppositions  as  to  the  relative  values  of  the  given 
numbers,  and  explaining  the  results.  We  will  illustrate  by 
the  following  example : 

Two  couriers  are  traveling  along  the  same  road,  in  the 
same  direction.     A  travels  m  miles  an  hour,  and  B  travels^ 
n  miles  an  hour.    At  12  o'clock  B  is  d  miles  in  advance  of  A. 
When  will  the  couriers  be  together  ? 

Suppose  they  will  be  together  x  hours  after  12.  Then  A  has  trav- 
eled mx  miles,  and  B  has  traveled  nx  miles,  and  as  A  has  traveled 
d  miles  more  than  B 


mx  —  nx  =  d.  .-.  x 


Discussion.    1.   If  m  is  greater  than  n,  the  value  of  x  is  positive, 
id  A  will  overtake  B  after  12  o'clock. 

2.  If  m  is  less  than  n,  the  value  of  x  is  negative.  In  this  case  B 
travels  faster  than  A,  and  as  he  is  d  miles  ahead  of  A  at  12  o'clock, 
A  cannot  overtake  B  after  12  o'clock,  but  B  passed  A  before  12 
o'clock.  The  supposition,  therefore,  that  the  couriers  are  together 
after  12  o'clock  is  incorrect,  and  the  negative  value  of  x  points  to  an 
error  in  the  supposition. 

3.  If  m  equals  n,  then  the  value  of  x  assumes  the  form  -  •    Now,  if 

the  couriers  are  d  miles  apart  at  12  o'clock,  and  if  they  travel  at  the 
same  rates,  it  is  obvious  that  they  never  will  be  together,  so  that  the 

symbol  -  may  be  regarded  as  the  symbol  of  impossibility. 

4.  If  m  equals  n  and  d  is  0,  then becomes  -  •     Now,  if  the 

*  '  m  —  n  0 

couriers  are  together* at  12  o'clock,  and  if  they  travel  at  the  same 
rates,  it  is  obvious  that  they  will  be  together  all  the  time,  so  that  x 

may  have  an  indefinite  number  of  values.    Hence,  the  symbol  -  may 

be  regarded  as  the  symbol  of  indetermination. 


204  PROBLEMS. 


Exercise  78. 


1.  A  train  traveling  b  miles  per  hour  is  m  hours  in 
advance  of  a  second  train  that  travels  a  miles  per  hour. 
In  how  many  hours  will  the  second  train  overtake  the  first  ? 

.           bm 
%  Ans.    r* 

a  —  o 

Discuss  the  problem  (1)  when  a  is  greater  than  b ;  (2)  when  a  is  equal 
to  b ;  (3)  when  a  is  less  than  b. 

2.  A  man  setting  out  on  a  journey  drove  at  the  rate  of 
a  miles  an  hour  to  the  nearest  railway  station,  distant  b 
miles  from  his  house.  On  arriving  at  the  station  he  found 
that  the  train  left  c  hours  before.  At  what  rate  per  hour 
should  he  have  driven  in  order  to  reach  the  station  just 
in  time  for  the  train  ? 

ab 


Ans. 


b 


Discuss  the  problem  (1)  when  c  =  0 ;  (2)  when  c  =  - ;    (3)  when 

c  = In  case  (2),  how  many  hours  did  the  man  have  to  drive 

from  his  house  to  the  station  ?    In  case  (3),  what  is  the  meaning  of 
the  negative  value  of  c  ? 

3.  A  wine  merchant  has  two  kinds  of  wine  which  he  sells, 
one  at  a  dollars,  and  the  other  at  b  dollars  per  gallon.  He 
wishes  to  make  a  mixture  of  I  gallons,  that  shall  cost  him 
on  the  average  m  dollars  a  gallon.  How  many  gallons 
must  he  take  of  each  ? 

Ans.   * ?-  of  the  first ;   ^ —/-  of  the  second. 

a  —  b  a  —  b 

Discuss  the  problem  (1)  when  a  =  b;  (2)  when  a  or  b  =  m;  (3) 
when  a  =  6  =  m ;  (4)  when  a  is  greater  than  b  and  less  than  m ;  (6) 
when  a  is  greater  than  b  and  6  is  greater  than  m. 


CHAPTER  XIII. 
SIMPLE   INDETERMINATE   EQUATIONS. 

199.  If  a  single  equation  is  given  with  two  unknown 
numbers,  and  no  other  condition  is  imposed,  the  number  of 
its  solutions  is  unlimited ;  for,  if  any  value  is  assigned  to 
one  of  the  unknown  numbers  a  corresponding  value  may  be 
found  for  the  other.  An  equation  that  has  an  indefinite 
number  of  solutions  is  said  to  be  indeterminate. 

200.  The  values  of  the  unknown  numbers  in  an  inde- 
terminate equation  are  dependent  upon  each  other  ;  so  that 
they  are  confined  to  a  particular  range. 

This  range  may  be  still  further  limited  by  requiring 
these  values  to  satisfy  some  given  condition;  as,  for 
instance,  that  they  shall  be  positive  integers. 

1.    Solve  Sx  +  Ay  =  22,  in  positive  integers. 

Transpose,  3  x  =  22  —  4y. 

Divide  by  3,  x  =  7  —  y +  1 


Transpose,  x  +  y  —  7  = 


3 

l-y 


1  —y 
Since  x  and  y  are  integers,  — r-2-  is  an  integer, 
o 

Let  — sr*  =  to,  an  integer. 

o 

Then,  y  =  1  —  3  to.  (1) 

Put  this  value  for  y  in  the  given  equation. 
Then,  x  =  6  +  4  m.  (2) 

In  equations  (1)  and  (2), 

If  to  =  0,  then  y  =  1  and  x  =  6. 

If  to  =  —  1,  then  y  =  4  and  x  =  2.     ' 

No  other  value  of  m  gives  positive  integers  for  both  x  and  y. 


206  SIMPLE  INDETERMINATE  EQUATIONS. 

2.    Solve  5  x  —  14  y  =  11,  in  positive  integers. 
Transpose,  5x  =  11  +  14  y. 

Divide  by  5,  x  =  2  +  2  y  +  *  *  **■ 

o 

1  +  4  w 
Transpose,  x  —  2  y  —  2  =  — ■*- —  • 

5 

1+42/ 

Then,  — - —  must  be  integral. 

T       1  +  4y            ._               5m  —  1        .       .      .     . 
Let  — - —  =  m,  then  y  = — >  a  fraction  in  form. 

To  avoid  this  difficulty,  it  is  necessary  to  make  the  coefficient  of  y 

equal  to  unity.     Since  — — -  is  integral,  any  multiple  of  — ■—-  is 
o  5 

integral.     Multiply  the  numerator  of  the  fraction,  then,  by  a  number 

that  will  make  the  division  of  the  coefficient  of  y  give  a  remainder  of  1. 

In  this  case,  multiply  by  4. 

We  have  i±J2*  =  g„  +  i±l. 

5  5 

4  +  y 
Let  — - —  =  ra,  an  integer, 

o 

Then,  y  =  5  ra  —  4.  (1) 

Since  x  =  I  (11  +  14?/),  from  the  original  equation, 

x  =  14  ra  —  9.  (2) 

Here  it  is  obvious  that  ra  may  have  any  positive  value. 

If  ra  =  1,  x  =    5,  y  =    1 

If  ra  =  2,  x  =  19,  y  =    6 

If  m  =  3,  a;  =  33,  y  =  11 

and  so  on. 

3.    Solve  5x  +  6  y  =  30,  so  that  sc  may  be  a  multiple  of  yf 
and  both  positive. 

Let  x  =  ray. 

Put  this  value  of  x  in  the  given  equation. 
Then,  (5  ra  +  6)  y  =  30. 

30 


.-.y  = 
and  x  = 


5ra  +  6' 
30  to 


5to  +  6 

Ifra  =  2,  x  =  3f,  y  =  l|; 

If  ra  =  3,  x  =  4fT  y  =  If 


SIMPLE  INDETERMINATE  EQUATIONS.  207 

Exercise  79. 
Solve  in  positive  integers : 

1.  2x  +  11  y  =  49.  5.    3x  +  8y  =  61. 

2.  7cc  +  3?/  =  40.  6.    $x  +  5y  =  97. 

3.  5x  +  7y  =  53.  7.    16aj  +  7y  =  110. 

4.  x  +  10y  =  29.  8.    7z  + 10^  =  206. 

Solve  in  least  positive  integers : 

9.    12a  -  ly  =  1.  12.  23ic-9y  =  929. 

10.  5x-17y  =  23.  13.  23x-33y  =  ±3. 

11.  23y-13a;  =  3.  14.  555 x  -  22 y  =  73. 

15.  A  man  spent  $114  in  buying  calves  at  $5  apiece, 
and  pigs  at  $3  apiece.     How  many  did  lie  buy  of  each  ? 

16.  In  bow  many  ways  can  a  man  pay  a  debt  of  $87 
with,  five-dollar  bills  and  two-dollar  bills  ? 

17.  Find  the  smallest  number  that,  when  divided  by  5 

or  when  divided  by  7,  gives  4  for  a  remainder. 

n  —  4  7t  ~~  4 

Let  n  =  the  number,  then  — - —  =  x,  and  — = —  =  y. 
o  7  . 

18.  A  farmer  sold  15  calves,  14  lambs,  and  13  pigs  for 
$200.  Some  days  after,  at  the  same  price  for  each  kind, 
he  sold  7  calves,  11  lambs,  and  16  pigs,  and  received  $141. 
What  was  the  price  of  each  ? 

First  eliminate  one  of  the  unknowns  from  the  two  equations. 

19.  A  number  is  expressed  by  three  digits.  The  sum  of 
the  digits  is  20.  If  16  is  subtracted  from  the  number  and 
the  remainder  divided  by  2,  the  digits  will  be  reversed. 
Find  the  number. 

20.  In  how  many  ways  may  100  be  divided  into  two 
parts,  one  of  which  shall  be  a  multiple  of  7  and  the  other 
a  multiple  of  9  ? 


CHAPTER    XIV. 
^EQUALITIES. 

201.  If  a  —  b  is  positive,  a  is  said  to  be  greater  than  b ; 
if  a  —  b  is  negative,  a  is  said  tp  be  less  than  b. 

Note.  Letters  in  this  chapter  are  understood  to  stand  for  positive 
numbers,  unless  the  contrary  is  expressly  stated. 

202.  An  Inequality  is  a  statement  in  symbols  that  one  of 
two  numbers  is  greater  than  or  less  than  the  other. 

203.  The  Sign  of  an  Inequality  is  >,  which  always  points 
toward  the  smaller  number.  Thus,  a  >  b  is  read  a  is 
greater  than  b ;    c  <  d  is  read  c  is  less  than  d. 

204.  The  expressions  that  precede  and  follow  the  sign  of 
an  inequality  are  called,  respectively,  the  first  and  second 
members  of  the  inequality. 

205.  Two  inequalities  are  said  to  subsist  in  the  same  sense 
if  the  signs  of  the  inequalities  point  in  the  same  direction ; 
and  two  inequalities  are  said  to  be  the  reverse  of  each  other 
if  the  signs  point  in  opposite  directions. 

Thus,  a  >  b  and  c  >  d  subsist  in  the  same  sense,  but  a  >  b  and  c  <  d 
are  the  reverse  of  each  other. 

206.  If  the  signs  of  all  the  terms  of  an  inequality  are 
changed,  the  inequality  is  reversed.  Thus,  if  a>^,  then 
—  a  <  —  b. 

207.  If  the  members  of  an  inequality  are  interchanged, 
the  inequality  is  reversed.     Thus,  if  a  >  b,  then  b  <  a. 


INEQUALITIES. 


209 


208.  An  inequality  will  continue  to  subsist  in  the  same 
sense  if  each  member  is  increased,  diminished,  multiplied,  or 
divided,  by  the  same  positive  number. 

Thus,   if    a  >  6,   then    a  +  c  >  6  +  c ;    a  —  c  >  6  —  c ;    oc>6c; 

aTC>6-rc.     Therefore, 


209.  A  term  can  be  transposed  from  one  member  of  an 
inequality  to  the  other,  provided  the  sign  of  the  term  is 
changed. 


Thus,  if  a  —  c>b, 

by  adding  c  to  both  members,  a  >  b  +  c. 


(§  208) 


210.  -4w  inequality  will  be  reversed  if  its  members  are 
subtracted  from  equal  numbers;  or  if  its  members  are 
multiplied  or  divided  by  the  same  negative  number. 

Thus,  iix  =  y  and  a > b,  then  x  —  a<y  —  b;  —  ac<  —  6c;  and 
a-f(-c)<&-f-(-c). 

211.  The  sum  or  product  of  the  corresponding  members 
of  two  inequalities  that  subsist  in  the  same  sense  is  an 
inequality  in  the  same  sense. 

Thus,  if  a  >  b  and  c  >  d,  then  a  +  c  >  6  +  eZ,  and  ac  >  bd. 


212.    The  difference  or  quotient  of  the   corresponding 
embers  of  two  inequalities  that  subsist  in  the  same  sense 

may  be  an  inequality  in  the  same  sense,  or  the  reverse,  or 

may  be  an  equality. 


Thus, 


7>4 


5>4 


By  subtraction,      -^  By  subtraction,      -^ 

4>2  J                      '2  =  2 

7>4  5>4 

By  division,         3>2  By  division,       -^^ 

2£>2  lf<2 


210  INEQUALITIES. 

1.    Find  one  limit  for  x,  if 


4sc 

3>   2       5 

Multiply  by  10, 
Transpose, 
Divide  by  25, 

40x- 

-30>15x- 
25  x  >  24. 

Find  the  limits  of 

X, 

x  - 

-4> 

2-3x, 

Sx- 

-2< 

a +  3. 

Transpose  in  (1), 
Divide  by  4, 
Transpose  in  (2), 
Divide  by  2, 

4x>6. 

x>H. 
2x<5. 

x<2|. 

6. 


given  a;  —  4  >  2  —  3  x,  (1) 

and  3a-2<a  +  3.  (2) 


Therefore,  the  value  of  x  lies  between  1^  and  2\. 

Exercise  80. 
Find  one  limit  for  x,  given : 

1.  (x  +  iy<x2  +  3x-5.  3.    x  +  2b>7x. 

4tX  —  2      S  —  5x  _         _      x  ,  _, 

2.  — 3— > 7 4-    3x-2<-  +  7£. 

5.  Find  the  limiting  values  of  x, 
given  4  x  —  6  <  2  #  +  4, 
and  2z  +  4>16-2a;. 

6.  If  a  <  ft,  find  the  limiting  values  of  #, 

ax  .   7  7       «2 

given  —  +  ox  —  a&  >  —  > 

,  to  b* 

and  y  —  a#  +  ab  <  y 

7.  Find  the  integral  value  of  x, 

given  J  (a  +  2)  +  £z  <•£-  (x  -  4)  -f  3, 

and  i  (x  +  2)  +  %x  >  £  (x  +  1)  +  i- 


INEQUALITIES.  211 

8.  Twice  a  certain  integral  number  increased  by  7  is  not 
greater  than  19 ;  and  three  times  the  number  diminished 
by  5  is  not  less  than  13.     Find  the  number. 

9.  Twice  the  number  of  pupils  in  a  certain  class  is  less 
than  3  times  the  number  minus  39 ;  and  4  times  the  num- 
ber plus  20  is  greater  than  5  times  the  number  minus  21. 
Find  the  number  of  pupils  in  the  class. 

213.    Theorem.     If  a  and  b  are  unequal,  a2  +  b2>2  ab. 

For  (a  —  b)2  must  be  positive,  whatever  the  values  of  a  aud  b. 
That  is,  (a  —  b)2  >  0. 

Squaring,  a2  -  2  ab  +  b2  >  0. 

Transposing  —  2  ab,  a2  +  b2  >  2  ab. 

If  a  and  b  are  positive  and  unequal,  show  that 
a*  +  b*>a2b  +  ab2. 

Now,  a3  +  63  >  a2b  +  ab2, 

if  (dividing  each  side  by  a  +  b) 

a2  —  ab  +  b2  >  ab, 
if  (transposing  —  ab)  a2  +  b2  >  2  a&. 

But  a2  +  b2  >  2  a&.  (Theorem) 

Therefore,  a3  4-  6s  >  a2b  4-  ao2. 

Exercise  81. 
If  the  letters  are  unequal  and  positive,  show  that : 

1.  a2  +  3b'2>2b  (a  +  b). 

2.  (a2  +  b2)  (a4  +  64)  >  (a8  +  J8)2. 

3.  a2b  +  a2c  +  a&2  +  &2c  +  ac2  +  be2  >  6  afo. 

4.  The  sum  of  any  fraction  and  its  reciprocal  >  2. 

5.  ab  +  ac  +bc  <  (a  +  ft  -  c)2+  (a  +  c-b)2+  (b  +  c-a)2 

6.  (a2  +  b2)  (e2  +  d2)  >  (ac  4-  6d)2. 

a  +  &        2ab  a        b       1,1 


CHAPTER   XV. 
INVOLUTION  AND  EVOLUTION. 


Involution. 

214.  The  operation  of  raising  an  expression  to  any  re- 
quired power  is  called  Involution. 

215.  Index  Law  for  Involution.     If  m  is  a  positive  integer, 

am  =  a  X  a  X  a to  m  factors. 

Consequently,  if  m  and  n  are  both  positive  integers, 
(an)m  =  an  X  an  X  an tow  factors 

—  qTi  +  n  +  n to  m  terms 

=  a7™.     Hence, 

Any  required  power  of  a  given  power  of  a  number  is  found 
by  multiplying  the  exponent  of  the  given  power  by  the  expo- 
nent of  the  required  power. 

216.  To  find  (ab)n. 
(ab)n  =  ab  X  cub to  n  factors 

=  (aX  a to  n  factors)  (b  X  b to  n  factors) 

=  anbn. 

In  like  manner,  (abc)n  =  anbnen\  and  so  on.     Hence, 
Any  required  power  of  a  product  is  found  by  taking  the 
product  of  its  factors  each  raised  to  the  required  power. 

217.  In  the  same  way  it  may  be  shown  that 
Any  required  power  of  a  fraction  is  found  by  taking  the 

required  power  of  the  numerator  and  of  the  denominator. 


INVOLUTION  AND  EVOLUTION.  213 

218.  From  the  Law  of  Signs  in  multiplication  it  is  evi- 
dent that  all  even  powers  of  a  number  are  positive ;  all  odd 
powers  of  a  number  have  the  same  sign  as  the  number  itself. 

Hence,  no  even  power  of  any  number  can  be  negative  ; 
and  the  even  powers  of  two  compound  expressions  that 
have  the  same  terms  with  opposite  signs  are  identical. 

Thus,  (b  -  af  =  (a-  b)\ 

219.  Binomials.     By  actual  multiplication  we  obtain, 

(a  +  iy^at  +  Zab  +  b2) 

(a  +  b)9  =  a8  +  3  a%  +  3  ab2  +  b* ; 

(a  +  by  =  a4  +  ±a*b  +  6a2b2  +  ±ab*  +  #. 

In  these  results  it  will  be  observed  that : 

1.  The  number  of  terms  is  greater  by  one  than  the  ex« 
ment  of  the  binomial. 

2.  In  the  first  term  the  exponent  of  a  is  the  same  as 
the  exponent  of  the  binomial,  and  the  exponent  of  a 
decreases  by  one  in  each  succeeding  term. 

3.  b  appears  in  the  second  term  with  1  for  an  exponent, 
id  its  exponent  increases  by  1  in  each  succeeding  term. 

4.  The  coefficient  of  the  first  term  is  1. 

5.  The  coefficient  of  the  second  term  is  the  same  as  the 
exponent  of  the  binomial. 

6.  The  coefficient  of  each  succeeding  term  is  found  from 
Le  next  preceding  term  by  multiplying  the  coefficient  of 
Lat  term  by  the  exponent  of  a  and  dividing  the  product 

)y  a  number  greater  by  one  than  the  exponent  of  b. 

220.  If  b  is  negative,  the  terms  in  which  the  odd  powers 
>f  b  occur  are  negative.     Thus, 


214  INVOLUTION  AND  EVOLUTION. 

1.  (a  -  b)s  =  a8  -  3  a%  -{-Sab2-  b*. 

2.  (a  -by  =  a4-4:  a8b  +  6a2b2  _4aS8  +  b\ 

By  the  above  rules  any  power  of  a  binomial  of  the  form 
a  ±  b  may  be  written  at  once. 

Note.    The  double  sign  ±  is  read  plus  or  minus  ;  and  a  ±  b  means 
a  +  b  or  a  —  b. 

221.  The  same  method  may  be  employed  when  the  terms 
of  a  binomial  have  coefficients  or  exponents. 

1.  Find  the  third  power  of  5  a;2  —  2y*. 
Since     (a  -  b)»  =  a*  -  3  a2b  +  3ab2-  b*, 

putting  5  x2  for  a,  and  2  y*  for  J,  we  have 
(5x2-2y*y 
=  (5a2)8  -  3  (5x*)2(2y*)  +  3  (5a2)  (2y8)2  -  (2y*)s 
=  125  x«  -  150  a?y  +  60  x2y«  -  8  f. 

2.  Find  the  fourth  power  of  x2  —  %  y. 

Since     (a  -  by  =  a4  -  4  a86  +  6  a262  -4ab*  +  b% 
putting  x2  for  a,  and  |  y  ? or  &»  we  nave 

=  (a2)4  -  4  (a2)8  (Jy)  +  6  (a5«)9(iy)»  -  4a2 (Jy)8  +  (Jy)4 
=  x9  -  2x*y  +  f  «y  -  }sY  +  ^y4. 

222.  In  like  manner,  &  polynomial  of  three  or  more  terms 
may  be  raised  to  any  power  by  enclosing  its  terms  in  paren- 

1  theses,  so  as  to  give  the  expression  the  form  of  a  binomial. 

1.    (a  +  b  +  cy  =  [a  +  (b  +  c)y 

=  a*  +  3a2(b+c)+3a(b  +  cy+(b+cy 

=  a9  +  3  a2b  +  3  a*c  +  3  ab2  +  6  afo 

+  3  ac2  +  b*  +  3  b2c  +  3  6c2  +  c«„ 


INVOLUTION  AND  EVOLUTION. 


215 


2.    (xs-  2x2-\-  3x  +  4y 

=  [(x3-2a;2)  +  (3a;  +  4)]2 

=  (xs  -  2  a:2)2  +  2  (x*  -2a:2)  (3x  +  4)  +  (3x  -f-  4)2 
=  a:6  -  4x5  +  4z4  +  6z4  -  4a;8  -  16x2  +  9x2  +  24a  +  16 
=  a:6- 4a;5  +  10a;4  -4a;8--  7 x2  +  24 a;  +  16. 


Exercise  82. 
Eaise  to  the  required  power : 


1.  (ay. 

2.  (a2b*)b. 

3'  \3ab*J 

4.  (-5a62c8)4. 

5.  {-lx2yz*)\ 
/_3aWY 

7.  (-2cc22/4)6. 

8.  (- 3  a268a;4)5. 

»■  (-sw- 


10.  (cc  +  2)6. 

11.  (a;2-2)4. 

12.  (x  +  3)\ 

13.  (2a;  +  l)6. 

14.  (2m2-l)5. 

15.  (2x  +  3yy. 

16.  (2  a;-?,)8. 

17.  (xy-2)\ 

18.  (1  -  a;  +  a;2)2. 

19.  (l-2a;  +  3a;2)3. 

20.  (1-a-fa2)8. 

21.  (3-4a;  +  5a;2)2. 


Evolution. 

223.  The  operation  of  finding  any  required  root  of  an 
expression  is  called  Evolution.  A  root  of  an  imperfect 
power  cannot  be  found  exactly. 

Thus,  the  exact  value  of  the  square  root  of  2  can  be  written  only 
as  V2,  and  the  exact  value  of  the  cube  root  of  4  can  be  written  only 

as  v4.     Approximate  values  of  these  expressions,  however,  can  be 
found  by  annexing  ciphers  and  extracting  the  root. 


1 


216  INVOLUTION  AND  EVOLUTION. 

224.    Index  Law  for  Evolution.     If  m  and  n  are  positive 
integers,  we  have 


(am)n  =  amn. 

(§215) 

Converselj, 
Also, 

Converselj, 
and 

n, ■* 

■y/amn  =  a  ■  =  am. 
(ab)n  =  an6w. 
y~tfW=  ^fan  X  V^"  = 
Va6  =  V^  X  J/b. 

(§  216) 

:  ab, 

Thus,  the  c^&e  n>o£  of  a6  is  a?  =  a2 ;  the  fourth  root  of 
81  a12  is  found  bj  taking  the  fourth  root  of  81  and  of  a12 ; 
and  is  3  a8.     Hence, 

225.  To  Find  the  Root  of  a  Simple  Expression, 

Take  the  required  root  of  the  numerical  coefficient,  and 
divide  the  exponent  of  each  letter  by  the  index  of  the  required 
root. 

226.  From  the  Law  of  Signs  it  is  evident  that : 

1.  Anj  even  root  of  a  positive  number  will  have  the  double 
sign,  ±. 

2.  There  can  be  no  even  root  of  a  negative  number. 

For  V—  x2  is  neither  +  x  nor  —  x ;  since  the  square  of  +  x  =  +  x2, 
and  the  square  of  —  x  =  +  x2. 

The  indicated  even  root  of  a  negative  number  is  call 
an  imaginary  number. 

3.  Anj  odd  root  of  a  number  will  have  the  same  sign  as 
the  number. 

©n      an 
—  ~Tn  •     (§  217) 

n  \nn         ~\l  tjft         n. 

Converselj,  a/—  =  -—  —  -r '         That  is, 

Any  required  root  of  a  fraction  is  found  by  taking  the 
required  root  of  the  numerator  and  of  the  denominator. 


INVOLUTION  AND  EVOLUTION. 


217 


„         ./16  a;2  4  a;       3/ 

Thus'  ^Slf  =  ±  %;    V-27mV  =  -  3 


mn2; 


*!l6xy2  2xhf 

\  81  a16   ~  ±   3  a4  ' 


227.  If  the  root  of  a  number  expressed  in  figures  is  not 
readily  detected,  it  may  be  found  by  resolving  the  number 
into  its  prime  factors.  Thus,  to  find  the  square  root  of 
3,415,104: 


28 

3415104 

28 

426888 

32 

53361 

7 

5929 

7 

847 

11 

121 

11 

3,415,104  =  26  X  32  X  72  X  ll2. 
V3,415,104  =  28  x  3   X7   X  11  =  1848. 


Simplify : 


i.  V4^y. 


2.    V64z9. 


3.    Vl6xy2 


4.    V-32a10. 


5.    V-27ica 


6.  V25a4. 

7.  v'-  8  a866. 


8.    V64*12. 


Exercise  83. 


9.    -v^-216 


10.    V729*18. 

6, 


11.    V243^10. 


12.  V-1728^8. 

13.  ■v'-  343  a\ 

14.  •^81^24. 

15.  -\/512a12616. 


16.    V*8Bi/12m. 


17,   V 


9a2^ 


18 


16ic42/2 

»<     8a?y 
\      27  *9 


20. 


21 


4 
■4 


S2a1Q 
243  x™ 


16  x4 

81a«b12' 


216  a24 


218  INVOLUTION  AND  EVOLUTION 


Square  Roots  of  Compound  Expressions. 

.    Since  the  square   of   a  -f-  b  is  a2  +  2  ab  +  b2,  the 
square  root  of  a*\+  2  ab  +  J2  is  a  4-  b. 

It  is  required  to  find  a  method  for  extracting  the  square 
root,  a  +  5,  when  a2  -\- 2  ab  -\- b2  is  given  : 

The  first  term,  a,  of  the  root  is  obviously  the  square  root  of  the 

first  term,  a2,  of  the  expression. 

a2  +  2 ab  +  b2 [a  +  6      If  the  a2  is  subtracted  from  the  given 

a2  expression,    the    remainder  is   2ab  +  b2. 

2a  +  b       20&  +  62  Therefore,  the  second  term,  &,  of  the  root 

2ab  +  b2  is  obtained  when  the  first  term  of  this 

remainder  is  divided  by  2  a,  that  is,  by 

double  the  part  of  the  root  already  found.    Also,  since 

2ab  +  b2  =  (2a  +  b)b, 

the  divisor  is  completed  by  adding  to  the  trial  divisor  the  new  term  of 
the  root. 

Find  the  square  root  of  25  x2  —  20  x*y  +  4  x*y*. 

25 x2  —  20 x*y  +  4 xiy2\5x  —  2x2y  ' 
25  x2 

—  20x*y  +  4x*y* 

—  20s8y  +  4s4y2 


10a;  —  2cc2y 


The  expression  is  arranged  according  to  the  ascending  powers  of  x. 

The  square  root  of  the  first  term  is  6 »,  and  5  x  is  placed  at  the 
right  of  the  given  expression,  for  the  first  term  of  the  root. 

The  second  term  of  the  root,  —  2x2y,  is  obtained  by  dividing 
—  20  xsy  by  10  x,  and  this  new  term  of  the  root  is  also  annexed  to  the 
divisor,  10  x,  to  complete  the  divisor. 

229.  The  same  method  will  apply  to  longer  expressions, 
if  care  is  taken  to  obtain  the  trial  divisor  at  each  stage  of 
the  process,  by  doubling  the  part  of  the  root  already  found, 
and  to  obtain  the  complete  divisor  by  annexing  the  new  term 
of  the  root  to  the  trial  divisor. 


INVOLUTION  AND  EVOLUTION 


219 


Find  the  square  root  of 

1  +  10a2  +  25x*  +  16a6  -  2±x5  -  20  a8  -4a. 

Arrange  the  expression  in  descending  powers  of  x. 

16x6-24x6  +  25x*-20x3  +  10x2-4x+114x3-3x2+2x-l 
16  x« 


$x*-~3x? 

-24x6  +  25^ 
-24x*  +    9x< 

—  20  x3  +  10x2 
-12x3+    4x2 

8x3- 

6x2  +  2x 

16x^ 
16x4 

8x8  —  6x2  +  4x 

—  1 

—  8x3  +    6x2-4x  +  l 

-  8x3  +    6x2-4x  +  l 

It  will  be  noticed  that  each  successive  trial  divisor  may  be  obtained 
by  taking  the  preceding  complete  divisor  with  its  last  term  doubled. 


Exercise  84. 
Find  the  square  root  of : 

1.   a4-8a8  +  18a2-8a  +  l. 


2.    9 


6  as  +  13  a2  -  4  a  +  4. 


3.  4:X*-12xsy  +  29xy-30xtf  +  25y\ 

4.  l  +  4a  +  10a2  +  12a8  +  9a4. 

6.  16-96a  +  216a2-216a8  +  81a4 

6.  a4-  22 xs  +  95 x2  +  286a  +  169. 

7.  4a4  -  11a2 +  25- 12a8  +  30a. 

8.  9a4 +  49 -12a8- 28a  +  46a2. 

9.  49  a4 +  126  a8 +  121 -73  a2 -198  a. 

10.  16a4-30a-31a2  +  24a8  +  25. 

11.  a4  -  2  ax*  +  3  a2a2  -  2  a8x  +  a\ 

12.  9a4- 18a +  1+87 a2- 54a8. 


220 


INVOLUTION  AND  EVOLUTION. 


230.  If  an  expression  contains  powers  and  reciprocals 
of  powers  of  the  same  letter,  the  order  of  arrangement 
in  descending  powers  of  the  letter  is  as  follows : 


•Ju  •    •£>•     J_« 


111 


Find  the  square  root  of 

9c2      4as2 
a2  +  9c2 


4x      101 

15  o       25 


Arrange  in  descending  powers  of  x. 

4x2 
9  c2 
i& 
9c2 


4x       101      6  c      9c2 
15  c       25       5x       x2 


2x 
3c 


6c 
5x 


1      3c 
5+  x 


4x     1 

4x 

101 
25 

1 
25 

3c      5 

15c  ' 
15  c 

4x 
3c' 

2      3c 
6       x 

4 

4 

6c      9c* 
5x      x2 
6  c      9c2 
5x+  x2 

231.   An  approximate  value  of  an  imperfect  square  can 
be  found  to  any  required  number  of  terms  as  follows ; 
Find  to  three  terms  the  square  root  of  x2  +px. 


x2+px 


x  + 


8x 


•t? 

px 

px 

*? 

2x4 

■p 

4 

*>2 
4 

8x 

INVOLUTION  AND  EVOLUTION.  221 

Exercise  85. 
Find  the  square  root  of  : 

1.  ±X*+-4:X*-}sX  +  ~ 

0    4a2  ,  462 

2.  -77- +  8+—- 

b2  a2 

n    „  ,  3  a2      a  ,    1 

4.   ^  +  -3-  +—-  +  J  +  -. 

tf      3j      41      3,       y* 
y*         y        lo      4a;      4r 

4  2  4 

7.    16x*  +  iJ-x2y  +  8x2  +  $  ya  +  ty  +  L 

9x*      Sx8      43<k2      7x      49 
4     .      2    +     4  2   +  4  ' 

9.    4a2  +  -^---ll  +  4a. 
or      a 

Find  to  three  terms  the  square  root  of : 

10.  a2  +  b.  13.    1  +  a.  16.  4  a2 +  3. 

11.  x2  +  \y.  14.    l-2a.  17.  4 -3a. 

12.  l  +  2a.  15.    4a2  +  26.  18.  4a2-  1. 


232.  Arithmetical  Square  Roots.  In  extracting  the  square 
root  of  an  arithmetical  number,  the  first  step  is  to  arrange 
the  figures  in  groups. 

Since  1  =  l2,  100  =  102,  10,000  =  1002,  and  so  on,  the 
square  root  of  a  number  between  1  and  100  lies  between  1 
and  10 ;  of  a  number  between  100  and  10,000  lies  between 


1 


222  INVOLUTION  AND  EVOLUTION. 

10  and  100.  In  other  words,  the  square  root  of  a  number 
expressed  by  one  or  two  figures  is  a  number  of  one  figure .; 
of  a  number  expressed  by  three  or  four  figures  is  a  num- 
ber of  two  figures ;  and  so  on. 

If,  therefore,  an  integral  square  number  is  divided  into 
groups  of  two  figures  each,  from  the  right  to  the  left,  the 
number  of  figures  in  the  root  will  be  equal  to  the  number 
of  groups  of  figures.  The  last  group  to  the  left  may  have 
one  figure  or  two  figures. 

Find  the  square  root  of  3249. 

32  49  (57  In  this  case,  a  in  the  typical  form  a2  +  2  ab  +  fc2 

25 represents  5  tens,  that  is,  50,  and  b  represents  7.     The 

107)  7  49  25  subtracted  is  really  2500,  that  is,  a2,  and  the  cora- 

7  49  plete  divisor  2a  +  6is2X50  +  7  =  107. 

233.  The  same  method  will  apply  to  numbers  of  more 
than  two  groups  of  figures  by  considering  a  in  the  typical 
form  to  represent  at  each  step  the  part  of  the  root  already 
found. 

It  must  be  observed  that  a  represents  so  many  tens  with 
respect  to  the  next  figure  of  the  root. 

Find  the  square  root  of  5,322,249. 

5  32  22  49(2307 

4 
43)132 

129 
4607)3  22  49 
3  22  49 

234.  If  the  square  root  of  a  number  has  decimal  places, 
the  number  itself  will  have  twice  as  many.  Thus,  if  0.21  is 
the  square  root  of  some  number,  this  number  will  be  (0.21)2 
=  0.21  X  0.21  =  0.0441 ;  and  if  0.111  is  the  root,  the  num- 
ber will  be  (0.111  )2  =  0.111  X  0.111  =  0.012321. 


INVOLUTION  AND  EVOLUTION. 


223 


Therefore,  the  number  of  decimal  places  in  every  square 
decimal  will  be  even,  and  the  number  of  decimal  places  in 
the  root  will  be  half  as  many  as  in  the  given  number  itself. 

Hence,  if  a  given  number  contains  a  decimal,  we  divide  the 
number  into  groups  of  two  figures  each,  by  beginning  at  the 
decimal  point  and  marking  toward  the  left  for  the  integral 
number,  and  toward  the  right  for  the  decimal.  We  must 
be  careful  to  have  the  last  group  on  the  right  of  the  deci- 
mal point  contain  two  figures,  annexing  a  cipher  when 
necessary. 

Find  the  square  root  of  41.2164  ;  of  965.9664. 

41.21  64  (6.42  9  65.96  64  (31.08 

36  9 

124)6  21  61)65 

4  96  61 


1282)25  64 
25  64 


6208)4  96  64 
4  96  64 


If  a  number  contains  an  odd  number  of  decimal 
places,  or  if  any  number  gives  a  remainder  when  as  many 
figures  in  the  root  have  been  obtained  as  the  given  number 
has  groups,  then  its  exact  square  root  cannot  be  found.  We 
may,  however,  approximate  to  its  exact  root  as  near  as  we 
please  by  annexing  ciphers  and  continuing  the  operation. 

The  square  root  of  a  common  fraction  whose  denominator 

is  not  a  perfect  square  can   be  found   approximately  by 

jducing  the  fraction  to  a  decimal  and  then  extracting  the 

)t ;  or  by  reducing  the  fraction  to  an  equivalent  fraction 

rhose  denominator  is  a  perfect  square,  and  extracting  the 

square  root  of  both  terms  of  the  fraction.     Thus, 

/(X625  =  0.79057 ; 


4 


VlO  _^  VlO  _  3.16227 
Vl6~     4  4 


=  0.79057. 


224 


INVOLUTION  AND  EVOLUTION. 


Find  the  square  root  of  3 ;   of  357.357. 


3.(1.732. 

1 
27)2  00 

189 
343) 11  00 
10  29 


3462) 71  00 
69  24 


3  57.35  70  (18.903. 

1 
28)2  57 

2  24 
369)33  35 
33  21 


37803)  14  7000 
11  3409 


Exercise  86 
Find  the  square  root  of : 

1.  289. 

2.  1225. 

3.  12,544. 

4.  253,009. 

5.  529,984. 


6.  150.0625. 

7.  118.1569. 

8.  172.3969. 


11.  640.343025. 

12.  100.240144. 

13.  316.021729. 


9.   5200.140544.     14.   454.585041. 
10.   1303.282201.     15.   5127.276025. 


Find  to  four  decimal  places  the  square  root  of  : 

16.  10.          19.    0.5.          22.    0.607.          25.    f.  28.  f 

17.  3.            20.    0.7.          23.    0.521.         26.    f.  29.  f. 

18.  5.            21.    0.9.          24.    0.687.          27.    £.  30.  fT. 


Cube  Roots  of  Compound.  Expressions. 

Since  the  cube  of  a  +  b  is  a9  4-  3  a2b  4-  3  ab2  +  bQ, 
the  cube  root  of 


a*  +  3  a2b  -f  3  ab2  +  b*  is  a  +  b. 


INVOLUTION  AND  EVOLUTION  225 

It  is  required  to  devise  a  method  for  extracting  the  cube 
root,  a  +  b.  when  a8  4-  3  a2b  4-  3  a&2  4-  &8  is  given  : 

The  first  term,  a,  of  the  root  is  obviously  the  cube  root  of  the  first 
term,  a8,  of  the  given  expression. 

a8  +  3a26  +  3  ad2  +  W\a  +  b 
8  a2  a8 


4  3  a&  +  62 


3a2  +  3a&  +  62 


3a26  +  3a&2  +  68 
3  a26  +  3  a&2  -f  &3 


If  a8  is  subtracted,  the  remainder  is  3  a?b  +  3  a&2  -f  &s .  therefore, 
the  second  term,  6,  of  the  root  is  obtained  by  dividing  the  first  term  of 
this  remainder  by  three  times  the  square  of  a. 

Also,  since  3  a%  +  3  a&2  +  &3  =  (3  a2  +  3  a&  +  52)  &?  the  complete 
divisor  is  obtained  by  adding  3  ab  4-  62  to  the  £ria£  divisor  3  a2. 

Find  the  cube  root  of  8  x*  4-  36  cc2y  4-  54  a^2  4-  27  y8. 

8x8  +  36x2y  +  54*y2  +  27y8j2x  +  3^ 
12x2  8x8 


(6x  +  3y)3y  =         +18xy  +  9y2 


12x2  +  18x2/  +  9  2/2 


36  x22/4-  64xy2  +  27^ 
36  x2y  + 54x2/2  + 27  2/8 


The  cube  root  of  the  first  term  is  2  x,  and  2  x  is  therefore  the  first 
term  of  the  root.     8  x8,  the  cube  of  2  x,  is  subtracted. 

The  second  term  of  the  root,  3  2/,  is  obtained  by  dividing  36  x2y  by 
8  (2  x)2  =  12  x2,  vhich  corresponds  to  3  a2  in  the  typical  form,  and  the 
divisor  is  completed  by  annexing  to  12  x2  the  expression 

{3(2x)  +  32/}32/  =  18x2/ +  92/2. 

237.  The  same  method  may  be  applied  to  longer  expres- 
jions  by  considering  a  in  the  typical  form  3  a2  4-  3  ab  4-  b2 
to  represent  at  each  stage  of  the  process  the  part  of  the  root 
already  found.  Thus,  if  the  part  of  the  root  already  found 
is  x  4-  y,  then  3  a2  of  the  typical  form  will  be  represented 
by  S(x  4-  yf\  and  if  the  third  term  of  the  root  is  4-  z,  the 
3  ab  4-  b2  will  be  represented  by  3 (as  4-  y)  z  4-  z2.  So  that 
the  complete  divisor,  3  a2  4-  3  ab  4-  b29  will  be  represented 
by  3(x  +  y)2  4-  S(x  4-  y)  z  4-  z\ 


226 


INVOLUTION  AND  EVOLUTION. 


Find  the  cube  root  of  xe  —  3  x5  +  5  xz  —  3  x  —  1. 

jx2-x-l 


3x* 


(3x2  _  X)(_  b)  =         -3x8+    xs 


3x4-3x3  +    x2 


x6-3x6  +  5x3-3x~l 
x6 

—  3x6  +  5x8 

—  3x5       +  3x4  —    x3 


3  (x2  -  x)2  =  3x4  -  6x8  +  3x2 
(3x2-3x- !)(-!)  = -3x2  +  3x  +  l 


3x4-6x8 


+  3x+l 


-3x4  +  6x8-3x-l 
-3x4  +  6x8  — 3x-l 


The  root  is  placed  above  the  given  expression  for  convenience  of 
arrangement  on  the  page. 

The  first  term  of  the  root,  x2,  is  obtained  by  taking  the  cube  root 
of  the  first  term  of  the  given  expression ;  and  the  first  trial  divisor, 
3  x4,  is  obtained  by  taking  three  times  the  square  of  this  term. 

The  first  complete  divisor  is  found  by  annexing  to  the  trial  divisor 
(3x2  —  x)  (—  x),  which  expression  corresponds  to  (3  a  +  b)  b  in  the 
typical  form. 

The  part  of  the  root  already  found  (a)  is  now  represented  by  x2  —  x; 
therefore,  3  a2  is  represented  by  3  (x2  —  x)2  =  3  x4  —  6  x8  +  3  x2,  the 
second  trial  divisor ;  and  (3  a  +  b)  b  by  (3  x2  —  3  x  —  1)  (—  1) ;  there- 
fore, in  the  second  complete  divisor,  3  a2  +  (3  a  +  b)  b  is  represented  by 
(3x4  -  6x8  +  3x2)  +  (3x2  -  3x  -  1)  (-  1)  =  3x4  —  6x8  +  3x  +  1. 


Exercise  87. 
Find  the  cube  root  of : 

1.  a*  +  3a2x  +  3ax2  +  x\ 

2.  S  +  12x  +  6x2  +  x\ 

3.  x6  -  3  ax5  +  5  a*x3  —  3a5x-  a\ 

4.  1  -  6x  +  21x2  -  Ux*  +  63a4  -  54aB  +  27a* 
6.    l-3x  +  6x2-7x*  +  6xi-3x*  +  x\ 

6.   »8-r-l-6a;-6a;64-15a;2-f-15ic4-20a;8. 


INVOLUTION  AND  EVOLUTION.  227 

7.  64a;6  -  144a:5  +  8  -  36a;  +  102  a;2  -  171  a;3  +  204a;4. 

8.  27  aG  -  27  a5  -  18  a4  +  17  a3  +  6  a2  -  3  a  -  1. 

9.  8  a;6  -  36a;5  +  66a;4  -  63a;3  +  33a;2-  9sc  +  1. 
10.  27  +  108a;  +  90a;2  -  80a;3  -  60a;4  +  48a;5  -8x\ 

a;9       3a;8       5a;6      3a;4      xs 

y\h  yU  y\1  ^10  ^9 

238.  Arithmetical  Cube  Roots.  In  extracting  the  cube  root 
of  an  arithmetical  number,  the  first  step  is  to  arrange  the 
figures  in  groups. 

Since  1  =  l3,  1000  =  103,  1,000,000  =  1008,  and  so  on,  it 
follows  that  the  cube  root  of  any  number  between  1  and 
1000,  that  is,  of  any  number  which  has'  one,  two,  or  three 
figures,  is  a  number  of  one  figure ;  and  that  the  cube  root 
of  any  number  between  1000  and  1,000,000,  that  is,  of  any 
number  which  has  fowr,  five,  or  six  figures,  is  a  number  of 
two  figures  ;  and  so  on. 

If,  therefore,  an  integral  cube  number  is  divided  into 
groups  of  three  figures  each,  from  right  to  left,  the  number 
of  figures  in  the  root  will  be  equal  to  the  number  of  groups. 
The  last  group  to  the  left  may  have  one,  two,  or  three 
figures. 

239.  If  the  cube  root  of  a  number  has  decimal  places, 
the  number  itself  will  have  three  times  as  many.  Thus, 
if  0.11  is  the  cube  root  of  a  number,  the  number  is 
0.11  X  0.11  X  0.11  =  0.001331.  Hence,  if  a  given  number 
contains  a  decimal,  we  divide  the  number  into  groups 
of  three  figures  each,  by  beginning  at  the  decimal  point 
and  marking  toward  the  left  for  the  integral  number,  and 


228 


INVOLUTION  AND  EVOLUTION. 


toward  the  right  for  the  decimal.  We  must  be  careful  to 
have  the  last  group  on  the  right  of  the  decimal  point  con- 
tain three  figures,  annexing  ciphers  when  necessary. 

240.    Notice  that  if  a  denotes  the  first  term,  and  b  the 
second  term  of  the  root,  the  first  complete  divisor  is 

3a*  +  3ab  +  b% 

and  the  second  trial  divisor  is  3  (a  -+-  by,  that  is, 

3a2  +  6ab  +  3b% 

which  may  be  obtained  by  adding  to  the  preceding  complete 
divisor  its  second  term  and  twice  its  third  term. 

Extract  the  cube  root  of  5  to  five  places  of  decimals. 


5.000(1.70997 
1 

3  X  102  =  300 

4000 

3(iOX  7)  =  210 

72=  49^) 
559  > 

3913 

259  J 

87  000  000 

3  X  17002  =  8670000 

3(1700X9)=  45900 

92  =     81 1 

8715981  >. 

78  443  829 

45981  J- 

8  556  1710 

3  X  17092  =  8762043 

7  885  8387 

670  33230 

613  34301 

After  the  first  two  figures  of  the  root  are  found,  the  next  trial 
divisor  is  obtained  by  bringing  down  the  sum  of  the  210  and  49 
obtained  in  completing  the  preceding  divisor  ;  then  adding  the  three 
numbers  connected  by  the  brace,  and  annexing  two  ciphers  to  the  result. 

The  last  two  figures  of  the  root  are  found  by  division.  The  rule  in 
such  cases  is  that  two  less  than  the  number  of  figures  already  obtained 
may  be  found  by  division  without  error,  the  divisor  being  three  times 
the  square  of  the  part  of  the  root  already  found. 


INVOLUTION  AND  EVOLUTION. 


229 


Exercise  88. 
Find  the  cube  root  of : 
.   4913.  3.    1,404,928. 


.5.   385,828.352. 
6.    1838.265625. 


2.    42,875.  4.    127,263,527. 

Find  to  four  decimal  places  the  cube  root  of : 
87.  9.    3.02.         11.    0.05.  13.    §.         16.    fT. 

.    10.         10.    2.05.         12.    0.677.  14.    f.         16.    ^. 

241.  Since  the  fourth  power  is  the  square  of  the  square, 
and  the  sixth  power  the  square  of  the  cube,  the  fourth  root 
Is  the  square  root  of  the  square  root,  and  the  sixth  root  is 
the  cube  root  of  the  square  root.  In  like  manner,  the 
eighth,  ninth,  twelfth,  root  may  be  found. 

Exercise  89. 
Find  the  fourth  root  of : 
81a4  +  108  a8  +  54  a;2  -f  12  a  4-  1. 
16x*-32ax*  +  24ta2x2-8a8x  +  a\ 
.    14- 4a  4- a8  4- 4a7  4- 10 a6  4- 16a8  4- 10 a2  4-  19a4  4- 16a5. 

Find  the  sixth  root  of : 
4.    l  +  6rf  +  d6  +  6d5  +  15^  +  20<28-f  15^. 
►.    729  -  1458  x  +  1215  a2  -  540  a8  -f  135  a4  -  18  a5  4-  a6. 
1  -  I82/  +  135?/2  -  540  y*  4-  1215 y4  -  1458  y5  +  729y6. 

Find  the  eighth  root  of: 
'.   1  -  Sy  4-  28i/2  -  56y*  4-  70y4  -  56y*  4-  28y«-  8y7  4-y8. 


CHAPTER  XVI. 
THEORY  OF  EXPONENTS. 

242.  If  n  is  a  positive  integer,  we  have  defined  an  to 
mean  the  product  obtained  by  taking  a  as  a  factor  n  times, 
and  called  an  the  nt\i power  of  a;  we  have  also  defined  -^/a 
as  a  number  which  taken  n  times  as  a  factor  gives  the  prod- 
uct a,  and  called  y/a  the  nth.  root  of  a. 

243.  By  this  definition  of  an  the  exponent  n  denotes 
simply  repetitions  of  a  as  a  factor;   and  such  expressions 

as  a  ,  a~8  have  no  meaning.  It  is  found  convenient,  how- 
ever, to  extend  the  meaning  of  an  to  include  fractional  and 
negative  values  of  n. 

244.  If  we  do  not  define  the  meaning  of  an  when  n  is 
fractional  or  negative,  but  require  that  the  meaning  of  an 
must  in  all  cases  be  such  that  the  fundamental  index  law 
shall  always  hold  true,  namely, 

am  X  an  =  am+» 


we  shall  find  that  this  condition  alone  will  be  sufficient  to 
define  the  meaning  of  an  for  all  cases. 


245.    Meaning  of  a  Zero  Exponent.     By  the  index  law, 


a0  X  an  =  a0  +  n  =  an. 


Cb 

Divide  by  an,  a°  =  —  =  1. 

Therefore,  the  zero  power   of  any  number   is  equal  to 
unity. 


THEORY   OF  EXPONENTS. 


231 


246.    Meaning  of  a  Fractional  Exponent.     By  the  index  law, 


*_„*+*_„§  _ 


a*  X  cr  =  a'    a  =  a3  =  a ; 

a§  X  a1  X  a*  =  al+i+4  =  a§  = 

af  X  a*  X  a1  X  a1 


af  +  f  +  f+l  =  a¥  =  a8 


-  -  i+- ton  terms  2 

an  X  an to  n  factors  =  an   n  =  an  =  a 


an  X  an  to  n  factors  =  an    n 

provided  ra  and  n  are  positive  integers. 

That  is,  a*  =  s/a ;  a*  =  -^a ; 


=  a  ■  =  a* 


.?_  i 


Va8;  «n=  V^ 


The  meaning,  therefore,  of  a",  where  ra  and  w  are  posi- 
tive integers,  is  the  nth  root  of  the  rath  power  of  a.     Hence, 
The  numerator  of  a  fractional  exponent  indicates  a  power 
md  the  denominator  a  root. 

247.   Meaning  of  a  Negative  Exponent.     By  the  index  law, 
f  n  is  a  positive  integer, 


But 


an  X  a-n  =  an-n  =  a°. 
a°  =  l. 

.'.  an  X  a~n  =  1. 


(§  245) 


That  is,  an  and  a~n  are  reciprocals  of  each  other  (§  169), 
that  a~u  =  — ,  and  an  = 


248.  Hence,  we  can  change  any  factor  from  the  numerator 
)f  a  fraction  to  the  denominator,  or  from  the  denominator 
to  the  numerator,  provided  we  change  the  sign  of  its  exponent. 

ab2  '  1 

Thus'  cW  may  be  written  a62c"  Sd~ 3'  or  a-i6-2c»d8  ■ 


232  THEORY  OF  EXPONENTS. 

249.  We  have  now  assigned  definite  meanings  to  frac- 
tional exponents  and  negative  exponents,  by  assuming  that 
the  index  law  for  multiplication,  am  X  an  =  am  +  n,  is  true  for 
all  values  of  the  exponents  m  and  n. 

It  remains  to  show  that  the  index  laws  established  for 
division,  involution,  and  evolution  apply  to  fractional  and 
negative  exponents. 

250.  Index  Law  of  Division  for  all  Values  of  m  and  n.  To 
divide  by  a  number  is  to  multiply  the  dividend  by  the 
reciprocal  of  the  divisor. 

Therefore,  for  all  values  of  m  and  n, 

Qtn  1 

-  =  a»X-  =  «»Xa-  (§217) 

=  am~n. 

251.  Index  Law  of  Involution  and  Evolution  for  all  Values 
of  222  and  22. 

To  prove  (am)n  =  amn  for  all  values  of  m  and  n. 

Case  1.  Let  m  have  any  value,  and  let  n  be  a  positive 
integer. 

Then,  (am)n  =  am  X  am  X  am to  n  factors 


=  a* 


i  +  m  +  m to  n  terms 


P 

Case  2.    Let  m  have  any  value,  and  n  =  -,p  and  q  being 
positive  integers. 

Then,  (am)«  =  y/(amy  =  vS 

mp 

=  aT. 

Case  3.    Let  m  have  any  value,  and  n  =  —  r,  r  being  a 
positive  integer  or  a  positive  fraction. 


THEORY   OF  EXPONENTS. 


(am)~r  = 


Then'  (a^      * 

Therefore,  (am)n  =  amn  for  all  values  of  m  and  n. 


233 
(§  247) 


252.    To  prove  (ab)n  =  anbn  for  any  value  of  n. 
Case  1.    Let  wbea  positive  integer. 
Then,  (ab)n  =  ab  X  ab  X  ab  to  n  factors 

=  (a  X  a ton  factors) (b  X  b to n  factors) 

=  anbn. 

P 
Case  2.   Let  n--,p  and  q  being  positive  integers. 

Then,  by  Case  1,  §  251,  since  q  is  a  positive  integer,  the  qth 
p            p            p 
power  of  (aby  =  (aby  X  (ab)* to  q  factors 

z    t\-  +  - too  term* 

=  (ab)i    « 

=  (aby  =  a*>b*>.  (By  Case  1) 


Also  by  Case  1,  §  251,  since  q  is  a  positive  integer,  the  ^th 
power  of 

p  p         p       p                                p       p 
a<ib*  =  (a«  X  a« to  q  factors)  (6«  X  6« to  q  factors) 

=  a*>bp. 

p 
But  the  qth.  power  of  (ab)*  =  (aby  =  a?b*. 

That  is,  [(«*)*]*  =  \_a^f. 

Extracting  the  ^th  root  of  each  member,  we  have 

p        p  p 
(ab)*  =  am. 

Case  3.    Let  n  =  —  r,r  being  a  positive  integer  or  fraction, 
1  1 


Then, 


(ab)- 


(ab)r      arbr 


—  a~rb' 


234 


THEORY  OF  EXPONENTS. 


EXAMPLES. 


1.   27-*  =  -^ 


27*       -^27      3 

2.  16f=(v/16)8  =  (±2)8  =  d=8. 

3.  a%  X  aTl  =  a1'1  =  a**  =  %. 

4.  a1  X  a*  X  a_i  =  a8**"*  =  a0  =  1. 


5.    (O 


6. 


i\6_  „-ix6 


•*-(-«)  =  a-3  +  s 


7.    </a*b-  3c~  4d  =  a%~  kT  W. 

8     f4a-V§= 1 = J =      1     =- 

k         }  (4a"»)*      4«a-»x*      8«"x      8 

/lBa-A-*.    /giy.V       27  6f       27  a8^1 
\  8168/  \16a~V 

10.    (3V3)- 


8a-f 


a'        a' 


(3sa-s)§      3§xfa~2      3*      ^ 

Exercise  90. 
Express  with  fractional  exponents : 

1.  ■&?.      3.    ■$€?.        5.    -v^5.      7.    V^  +  ■%  4-  ■v/16« 

2.  Vol       4.    ^—8.     6.    "V/oi,       8.    VaV  +  Va8^. 


THEORY   OF  EXPONENTS.  235 


Express  with  radical  signs  : 
9.    a*.  11.    ah*.  13.    x*y~*.  15-    a*  -  xh*. 

10.    c*  12.    a*£*.  14.    3aj*y~*.       16.    a*  +  »V. 


Express  with  positive  exponents : 

2a-V 
17.    a~8.         19.    3x~Y-     21.    4z-y-2.      23. 


3-2^2^-6 


18.    a~*.         20.    4cc?/-7.       22.    3  a"4ft* 


Write  without  denominators : 

24-  -6  •  26-  — ££=•? 

as-6  a-Joc  2a 


25.    — : — r  27. 


28. 

X 

a~ 

-2£-f 

29. 

a" 

a" 

40. 

dW" 

1  x  (A)* 

-2^-2c-4 

Find  the  value  of  : 

30.  8*.  35.    (-27)1 

31.  16"*.  36.    (-  27)*  X  25*.  41.  flf"M  X  ah*. 

32.  27*.  37.    81_l  X  16*.  42.  (a~k*)~K 

33.  (-8)"*.         38.    8}X44.  43.  (cTV1)-2. 

34.  36f.  39.    (^V)^Xl6~l  44.  (»*y"***)H 

If  a  =  &,  b  =  2,  c  =  1,  find  the  numerical  value  of : 

45.  aV1.       47.    a~^2.         49.    3  (abf.         51.    (a£2c)*. 

46.  a£-2-         48.  a~V~*         50.    2  (a&)~*.       52.    (ab2c)~*. 


236  THEORY  OF  EXPONENTS. 

Multiplication. 
Multiply  a?*  +  as*y*  +  y*  by  «*  -  » V  4-  y* 
«*  +  &M  +  y* 

x  +  x*yi  +  »*y* 

+  a?*y*  +  g*y*  +  y 

a;  +  x  V  +  y 

Exercise  91. 
Multiply : 

1.  a*  +  J*  by  a*  -  &*  3.    a*  -  ft*  by  a*  -  b*. 

2.  a*  +  6*  by  a*  +  6*.  4.    a;1  +  2*  by  x1  -  2  a. 

5.  x~2  +  £C_1y" *  4-  y-2  by  a;-2  —  ar^y1  +  y~\ 

6.  a;*  —  aj*y*  +  y*  by  a;*  -f  y* 

7.  «*  +  ojV  +  y*  by  »*  -  y* 

8.  l  +  a-i  +  a-ityi^a-i+fr-t 

9.    aV*  +  2a*-3ft*by  2*"*- 4«T*- 6«rV. 
Division. 

Divide  Vaj1  +  Va;  -  12  by  v£  -  3. 
xi+    x*-12|x*-3 


«*  — 3**  x*  +  4 


+  4x*-12 
+  4x*-12 


r 

Hereaj*  +  ^  =  **-**=«*;  4a;8  +  a;8  =  4  a:'"  *  =  4a*° 


THEORY   OF  EXPONENTS. 


237 


Exercise  92. 


3.  a  —  b  by  a*  —  b  . 

4.  a  +  Jbyc^  +  6*. 


Divide : 

1.  a-b  by  a*  —  fr* 

2.  a  +  #  by  a*  +  **. 

5.  aj-3aj*  +  3a;*-l  by  »*  - 1 

6.  a;  -f  a?  V  +  ^  by  x       *  y   +  y  • 

7.  aj*  -  4sc*  4- 1  4-  6aT*  by  a£  -  2. 

8.  9x-12x*  -2  +  ±x~*  +  x-1  by  3  a;*  -  2  -  a;""*. 

9.  2ar-2  +  6ar-yi-16ar52r4  by  2  a;  4- 2  a:2?/- x  4- 4  a;  V* 

ill  i         i         l 

10.   x  +  y  4-  s  —  3  xJyJzJ  by  as*  4-  y   +  *  . 

Square  Boot. 
Find  the  square  root  of 

9  a;-4  -  18  aj~«y*  +  15  «T>  ~  6  x~  V  +  2/2- 
9a._4 — I8x-  «y*  4-  15x-2y  —  6  a- ty*  +y2  |3x-2— 3^-iyi+y 


9x-* 


—lSx-tyt  +  lbx-tp 
-18x-^4-  9x-2y 


6  s-2  —  6a- ^ +2/ 


6x-2y_  6x-!^4-y2 
6x~2y— 6a-V+y2 


Exercise  93. 
Find  the  square  root  of  : 

1.   a?*4-2aj*4-i 


4**4-4. 


2.    4a*-4a*J*4-J* 


3.  a; 

4.  4  a-2  4-  4  a'1  4-1. 

5.  9a -12a*  4- 10 -4a"*  4- a"1. 

6.  m24-2m  — l-2m-14-m-2. 

7.  49^-28x4-18^-4^4-1. 


CHAPTER   XVII. 
RADICAL  EXPRESSIONS. 

253.  A  radical  expression  is  an  expression  affected  by 
the  radical  sign ;  as  Va,  V5,  w,  Va~+~b,  V32. 

254.  An  indicated  root  that  cannot  be  exactly  obtained 
is  called  a  surd.  An  indicated  root  that  can  be  exactly 
obtained  is  said  to  have  the  form  of  a  surd. 

The  required  root  shows  the  order  of  a  surd ;  and  surds 
are  named  quadratic,  cubic,  biquadratic,  according  as  the 
second,  third,  or  fourth  roots  are  required. 

The  product  of  a  rational  factor  and  a  surd  factor  is 
called  a  mixed  surd;  as  3V2,  b^fit.  The  rational  factor 
of  a  mixed  surd  is  called  the  coefficient  of  the  surd. 

When  there  is  no  rational  factor  outside  of  the  radical, 
sign,  that  is,  when  the  coefficient  is  1,  the  surd  is  said  to  be 
entire;  as  V2,  Va. 

255.  A  surd  is  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  integral  and  as  small  as  possible. 

Surds  are  said  to  be  similar  if  they  have  the  same  surd 
factor  when  reduced  to  the  simplest  form. 

Note.  In  operations  with  surds,  arithmetical  numbers  contained 
in  the  surds  should  be  expressed  in  their  prime  factors. 

Reduction  of  Radicals. 

256.  To  reduce  a  radical  is  to  change  its  form  without 
changing  its  value. 


RADICALS.  239 

Case  1. 

257.   When  the  radical  is  a  perfect  power  and  has  for  an 
exponent  a  factor  of  the  index  of  the  root. 

1.  ■fa~*  =  J  =  a*  =  Va. 

2.  -y/MtfP  =  ^(6  aby  =  (6  ah)1  =  (6  abf  =  V6~a~b. 

3.  ~Z/25  JW  =  ^(6aa*c4)2  =  (5  a^c4)1  =  (5  a2^4)* 

=  -v/5^2^4. 
We  have,  therefore,  the  following  rule : 
Divide  the  exponent  of  the  power  by  the  index  of  the  root. 


Exercise  94. 

Simplify : 

l.    ^25. 

6.    -yfiM*. 

2.    ^16. 

7.    -v^5. 

3.    ^27. 

8.    -\/M\ 

•  4.    y/t&. 

9.    -f/27  a8b6. 

5.    -^64. 

10.    "^16a464. 

11. 


e  /25  a2 
^6462' 


Af(as 


12.  16^ 


3)5 


Case  2. 


^ 


13.       '  ^ 


8scy 


258.    When  the  radical  is  the  product  of  two  factors,  one  of 
which  is  a  perfect  power  of  the  same  degree  as  the  radical. 

Since  -\/aFb  =  Van  X\/J  =  (fv/J(§  224),  we  have 

1.  Va%  =  Va*xVb==aVb] 

2.  -^108  =  ^27  X  4  =  -^27  X  -y/i  =  3^4 ; 


240  RADICALS. 

3.  4  V72 a2b*  =  4  V36 a%2  X2b  =  4  V36 a%2  X  V2b 

=  4X  6oiV2J  =  24aiV26; 

4.  2^54 a46  =  2^27 a*x2ab  =  2^/2T~a~8  X  V^S    ' 

=  2  X  Sa-^/2ao  =  6a^/2aJ. 
We  have,  therefore,  the  following  rule : 

Resolve  the  radical  into  two  factors,  one  of  which  is  the 
greatest  perfect  power  of  the  same  degree  as  the  radical. 

Remove  this  factor  from  under  the  radical  sign,  extract 
the  required  root,  and  multiply  the  coefficient  of  the  surd  by 
the  root  obtained. 


Exercise  95. 


Simplify : 


1.  V28.  13.  7V144.  25     it  Uxy  - 

r-  . V27m8n* 

2.  V72.  14.  8VwA. 

3.  </72.  15.  3</bW.  26.    ^p; 

4.  -v^OO.  16.  2^V.  . 

8j - 27      »/mg 

5.  V432.  17.  11  -y/a12b16.  '    \'216^S' 


3/ 


6.  Vl92.  18.    7V8^.  ^    2«/^ 

7.  -\/l28.  19.    6^27  mV.  ^243 


8.  "^243.  20.  4-V^Sy.  29     *EZ.. 

, 8/ VlMfi 

9.  V176.  21.  V1029. 

10.  V405.  22.  ^2187.  30.   jj&l 

11.  2^112.  23.  "^1250. 

12.  3^/S6l.  24.  4^648. 


1296 
"612 

3  ah 
31. 


Sab    I  20c2 
2c  \9a2b2' 


RADICALS. 


241 


Case  3. 

259.  When  the  radical  expression  is  a  fraction,  the  denomi- 
nator of  which  is  not  a  perfect  power  of  the  same  degree  as  the 
radical. 


V|  =  V|  =  Viox^  =  iVIo. 


2  VA-V^-^-v^-*v* 

3-  <§=^^^r4=4 


5X3X4 

27  X  8 
1         3 


60  X 


27  X  8 


3x2 


60  =  J  V60. 


We  have,  therefore,  the  following  rule : 

Multiply  both  terms  of  the  fraction  by  a  number  that 
will  make  the  denominator  a  perfect  power  of  the  same 
degree  as  the  radical ;  and  then  proceed  as  in  Case  2. 


Exercise  96. 

Simplify : 

i.  2Vf 

4.    7VJ:             7.    -v^f. 

10. 

2^|. 

2.    fVf. 

4/ 3/- 

5.     Vff               8.    Vf. 

11. 

3^A- 

3-    iVj. 

6.    3V^.            9.    %/fc 

12. 

2^£ 

i3-Vf 

15.    \S£t .            17. 

4 

>2cy2 
6d2 

14.    &. 

16.    J/JS-.         18. 

21 

/2  a862c 

'yse 


242  RADICALS. 

Case  4. 

260.  To  reduce  a  mixed  surd  to  an  entire  surd. 

Since  a^/b  =  Va"  X  Vd  =  "VaFb,  we  have 

1.  3VS  =  V32~x~5  =  V9~X~5  =  V45; 

2.  a% Vbc"  =  V(a26)2  X  be  =  s/aW  X  be  =  VaWc ; 

3.  2a-v/cciy=:^(2a;)8  X  ^  =  "V/8cc8  X  xy  =  J/S x*y ; 

4.  3  y2^  =  *t/(3  y2Y  X  x8  =  ^81  y8^8. 
We  have,  therefore,  the  following  rule : 

liaise  the  coefficient  to  a  power  of  the  same  degree  as  the 
radical,  multiply  this  power  by  the  given  surd  factory  and. 
indicate  the  required  root  of  the  product.  f 

Exercise  97. 
Express  as  an  entire  surd : 

1.  5VE.  5.  2^3.  9.  -2Vy.  13.  iVa. 

2.  3Vll.  6.  3^2.  10.  -3VsA  14.  -JVa2. 

3.  3^3.  7.  2-Z/2.  11.  -m-v^O.  15.  |V»T8. 

4.  2^4.  8.  2^4.  12.  -2-s/x.  16.  -  i^/m\ 

Case  5. 

261.  To  reduce  radicals  to  a  common  index. 

Keduce  V2  and  V3  to  a  common  index. 

V2  =  2*  =  28  =  v^»  =  V8. 

V3  =  3*  =  38  =  V32  =  V9. 
We  have,  therefore,  the  following  rule : 


RADICALS.  243 

Write  the  radicals  with  fractional  exponents,  and  change 
these  fractional  exponents  to  equivalent  exponents  having  the 
least  common  denominator.  Raise  each  radical  to  the  power 
denoted  by  the  numerator,  and  indicate  the  root  denoted  by 
the  common  denominator. 

Exercise  98. 
Eeduce  to  surds  of  the  same  order : 


I. 

■^3  and  y/B. 

7. 

V2,  ^3,  and  yfc . 

2. 

vTIand  V6. 

8. 

Va2,  Vft,  and  Vc. 

3. 

V2  and  y/Z. 

9. 

Vo*,  Vc®,  and  Vcc8. 

4. 

Va-  and  V#~2. 

10. 

y/x2y,  Vak,  and  "v2». 

5. 

V5  and  VfE. 

11. 

6/ 4/ 

Vsc  —  y  and  Vsc  +  y. 

3. 

2*  2f ,  and  2* 

12. 

3/ I 

Va  +  b  and  V a  —  6. 

Note.  Surds  of  different  orders  may  be  reduced  to  surds  of  the 
same  order  and  then  compared  in  respect  to  magnitude. 

Arrange  in  order  of  magnitude : 

13.  -v^lS  and  V6.  15.    V^O,  V^,  and  V& 

14.  yfl  and  V§.  16.    V3,  y%,  and  yft. 

Addition  and  Subtraction  of  Radicals. 

262.  In  the  addition  of  surds,  each  surd  must  be  reduced 
to  its  simplest  form ;  and,  if  the  resulting  surds  are  similar, 

Find  the  algebraic  sum  of  the  coefficients,  and  to  this  sum 
annex  the  common  surd  factor. 

If  the  resulting  surds  are  not  similar, 

Connect  them  with  their  proper  signs. 


244  RADICALS. 

1.  Simplify  V27  +  V48  +  Vl47. 

V27  =  (32  x  3)*  =  3  X  3*  =  3 V3; 

V48  =  (2*  X  3)*  =  22  X  3*  =  4  X  3*  =  4  V3; 

Vl47  =  (72  X  3)*  =  7  X  3*  =  7  V3. 

V27  +  V48  +  Vl47  =  (3  +  4  +  7)  V3  =  14  V3. 

2.  Simplify  2^320  -3  -^40. 

2V320  =  2(2°  X  5)*  =  2  X  22  X  5*  =  sVS; 
3V4O  =  3(28  X  5)*  =  3  X  2  X  5*  =  6 VS. 
.-.  2  V320  —  3  ViO  =  (8  -  6)  V6  =  2  VS 

3.  Simplify  2V£- 3  Vf  +  V^. 

2 Vf  =  2 V^  =  2V15  X  I  =  f Vl5; 
3V|  =  3 V||  =  3 V15  X  £  =  I Vl5;      . 

.•.2V|-3V|+VS=(^-3+T25)Vl6  =  |Vl6. 

Exercise  99. 
Simplify  : 

1.  4VlT  +  3VTT-5VlT. 

2.  2V3 -5V3  +  9V3. 

3.  5"V/4  + 2^32 --5^108.  7.  V27  +  V48  +  V75. 

4.  3^2  +  4-^2— "^64.  8.  4Vl47-f  3V75  -f  VT92. 

5.  ^-5/5  +  2^-v/5  +  iV/40.  9.  V^  +  |V^-ffV^. 

6.  3^5  —  5^+  -^243.  10.  -V^  +  i^-S-J/^T^. 


RADICALS. 


24:5 


11.  Va*  +  bVa  -  3Va. 

12.  \r25b  +  2V9b-3VIb. 

13.  2VI75-3V63  +  5V28. 

14.  V2  +  3V32  +  ^Vl28-6Vl8. 

15.  V75  +  Vi8  -  Vl47  +  V300. 

16.  20V245- V5  + Vl25-2^Vl80. 

17.  2V20  +  ^Vl2-2V27  4-5V45-9Vl2. 

18.  7V25  +  4V45- V9-2V80  + V20-4VSI. 

19.  -#54  +  V£- V^-fVf. 

20.  2V|  +  V60-  Vl5  +  Vf  +  V^. 

21.  ■v/27V  -  -#8^  +  -^125^. 

22.  Vo^-V^+^32J. 

23.  Va4^  +  V^  -  V4a%. 

24.  V4  ic22/2^  +  Vy4^  4-  Va^s. 

25.  VaWc  -  aVIc  +  bVa?c. 

26.  -#81  a5  -  ^16^  +  ^256  a5. 

27.  "#27  m4  -  -#125  m  +  "#216  m. 

28.  V8^  -  V50  a8  -  3  VlSU. 


29.    6  a  V63  a&8  -  3  V112  a%z  +  2ab  V343  a*. 


30.    3  V125  mV  +  rc  V20  m8  -  V500  m8^2. 


31.  V32  a465  -f  6  Vm  +  3  V128  a?b\ 

32.  2-J^  -  3a2^64l  -f5ftvS  +  2a2^1255. 


246  RADICALS. 

Multiplication  of  Radicals. 
263.    Since  Va  X  Vb  =  Vab,  we  have 

1.  3V8  X  5V2  =  3  X5  X  V8  X  V2  =  15  Vl6  =  60 ; 

2.  3V2  X  4-^  =  3 vT8X  4^=12^72. 

We  have,  therefore,  the  following  rule : 

Express  the  radicals  with  a  common  index.  Find  the 
product  of  the  coefficients  for  the  required  coefficient,  and  the 
product  of  the  surd  factors  for  the  required  surd  factor. 

Reduce  the  result  to  its  simplest  form. 

Exercise  100. 
Find  the  product  of : 

1.  V3  X  V27.  7.  vlxv^.  13.  -^54  X  V& 

2.  V5xV20.  8.  ^27X^9.  14.  2  V8  X  VS. 

3.  VS  X  Vl8.  9.  VS  X  Vl2.  15.  a/8  X  V^4. 

4.  V3  X  -vft  10.  V3  X  VS.  16.  ^7  X  V-49. 

5.  ^2x^32.  11.  V^X-v^.  17.  ^/SlX^Id. 

6.  ^27X^3.  12.  ^6x^8.  18.  §^18X1^3. 

19.  (V18  +  2V72-3V8)  X  VS. 

20.  (V^2-^V864  +  3^4)  X  $2. 

21.  (J  V27  -  J  V2187  +  }  V432)  X  V3. 

22.  VH  X  y/t.  25.    V3  X  -\/72.        28.    V81  X  V& 

23.  VT6  X  V250.    26.    Vf  X  -v^.  29.    V^  X  Vf . 

24.  V"64  X  ^lfr      27.    V^  X  V^.        30.     Vf  X  Vj. 


RADICALS.  247 

264.    Compound  radicals  are  multiplied  as  follows  : 

Multiply  2  V3  +  3  Vx  by  3  V3  -  4  Vac. 

2V3  +  3VX 
3V3-4Vx 


18  +  9V3x 

—  8Vsx  —  12* 

18+    V3x-12x 
Exercise  101. 


Multiply 


1.  V5  +  ^*  by  V5  _  ^      4-    8  +  3V2by  2-  V2. 

2.  V9  -  ^  hY  V9  +  'v/^-  5-    5  +  2V3by3-5V3. 

3.  3  +  2V5  by  2  -  VS.  6.   3  -  V6  by  6  -  3 Vo\ 

7.  2V6-3V5  by  V3  +  2V2. 

8.  7  -  V3  by  V2  +  V5. 

9.  V/9-2v/4by  4-^3+^2. 

LO.  2V30-3V5  +  5V3  by  V8  +  V3-  VB. 

.1.  3V5-2V3  +  4V7  by  3V7-4VS-5V3. 

L2.  4 V8  +  i Vl2  -  i V32  by  8  V32  -  4V50  -  2 V2. 

L3.  ^6-^3  +  ^16  by  ^36  +  ^9-^4. 

L4.  2Vf-8Vf +  3V|  by  3Vf- Vl2- V6. 

L5.  2V|-4V|-7Vf  by  3V|-5V30-2VY. 

L6.  2Vl2  +  3V3  +  6Vi  by  2Vl2  +  3V3  +  6V£. 


248  RADICALS 


Division  of  Radicals. 

-Vab       Va  X  -Vb        nr- 
265.    Since  — —  = =  Wb,  we  nave 

2V2 
2  4^3  =  4^3*  =  4^32  x  23  _  t^ 

2V2      2W         2-v^ 

We  have,  therefore,  the  following  rule : 

Express  the  radicals  with  a  common  index.      Find  the 

quotient  of  the  coefficients  for  the  required  coefficient,  and  the 

quotient  of  the  surd  factors  for  the  required  surd  factor. 

Reduce  the  result  to  its  simplest  form. 

Exercise  102. 
Divide : 

1.  V243  by  VS.        4.   V£  by  Vf        7.   V}$  by  Vffj 

2.  -v^T  by  VS.  5.   V#  by  V§.         8.   Vjf  by  V^U 

3.  VFa7  by  Va5.       6.   V^  by  Vf       9.   Vjf  by  V$£, 

10.  3V6  +  45V2  by  3V3. 

11.  42V5-30V3  by  2Vl5. 

12.  84V15  +  168V6  by  3V21. 

13.  30-5^4  —  36^10  +  30^90  by  3V^0. 

14.  50VT8  +  184/20-48^/5  by  2^30. 

15.  V54  by  "^36.     17.   VT2  by  Vo*.     19.   Vf  by  ^5|. 

16.  -V^49  by  sfl.       18.   V^  by  "V^.  20.   V^a-  by  Vx*. 
21.   VU064  by  VlO.         22.   V»a  -  y3  by  x  +  yr 


RADICALS.  249 

The  quotient  of  one  surd  divided  by  another  may 
be  found  by  rationalizing  the  divisor ;  that  is,  by  multiply- 
ing the  dividend  and  divisor  by  a  factor  that  will  free  the 
divisor  from  surds.  This  method  is  of  great  utility  when 
we  wish  to  find  the  approximate  numerical  value  of  the 
quotient  of  two  simple  surds,  and  is  the  method  required 
when  the  divisor  is  a  compound  surd. 

1.  Divide  3  V8  by  V5. 

3V8      6V2_6V2  X  V6_6Vl2         r-  _ 

2.  Divide  3VS  -  4  V2  by  2V5  +  3V2. 

Multiply  the  dividend  and  the  divisor  by  2V5  —  3  V2, 

(3V5  -  4  V2)  (2  V5  -  3V2)  _  54  -  17  VlO 
(2V6  +  3V2)(2V6-3V2)     *    20-18 

64-17V10      n„       17    rrz       „ 
= - =  27  —  V-  v  10.     Hence, 

267.   When  the  divisor  is  a  binomial  containing  surds  of  the 
second  order  only, 

Multiply  the  dividend  and  the  divisor  by  the  divisor,  with 
he  sign  between  the  terms  changed. 

Exercise  103. 
Divide : 

Va  +  V&  by  Vab.  7.   3  +  5 Vf  by  3  -  5V7. 

a.    Vl25by5V65.  8.   21 V3  by  4a/3  -  3V2. 

3.  3  by  11  + 3 V7.  9.    75VII  by  8V2  +  2Vt. 

4.  3V2-lby3V2  +  l.  10.  V5  -  Vs  by  VE  +  V3. 
6.  17  by  3 V7  +  2  V§.  11.  V8  +  Vf  by  V7  -  V2. 
6,   lby\/2  +  V3.  12.   7-3Vl0by  5  +  4V5. 


250  RADICALS. 

Given    V2  =  1.41421,     V§  =  1.73205,  V5  =  2.23607; 

find  to  four  places  of  decimals  the  value  of : 

13.  i°        16.     i    .      19.     i  .  22.  7-3VB. 

V2  V500  3V2  5  +  4V5 

l4. 4.    17;    1  .    20.    1  .  23.  iws. 

V3  V243  V125  V5-2 

-c    12         ,0      l  1  «     3V2-1 

15.    -t=«  18.    =•  21.    =*•  24.    = 

V5  2V3  4V5  3V2  +  1 


Involution  and  Evolution  of  Radicals. 

268.   Any  power  or  root  of  a  radical  is  easily  found  by 
using  fractional  exponents. 

1.  Find  the  square  of  2  Va. 

(2  Va)2  =  (2  a*)2  =  22  as  =  4  a*  =  4  Va2. 

2.  Find  the  cube  of  2  Va. 

(2  Va)8  =  (2  a*)8  =  23  a*  =  8  a*  =  8  a  Va. 

3.  Find  the  square  root  of  4  a?  ~Va8bs. 

(4  a  Vo3^)*  =  (4  xa*&$)*  =  4*  xM&*  =  4*  xMfc*  =  2  V^x2. 

4.  Find  the  cube  root  of  4ccVa8&8. 

(4x  Va8^)*  =  (4  xo*6*)*  =  4Ma*6*  =  **a&W  =  Vl6  a^x2. 

Exercise  104. 
Perform  the  operations  indicated : 
1.    (^T2)8.  3.    ("Vx*)15.  5.    VV(x-y)8. 


2.    (Vm5)6.  4.    (Vy*)12.  6.    V-V7^-^)80. 


RADICALS.  251 

7.  (V2^b)\  10.    V^.  13.    V^(3a-2^. 

8.  (^2  _  yy,       11#   V-^729.         14.   V^32a46. 

9.  (Vz)4.  12.  VV125.  15.   Vl28v'243a^. 

Properties  of  Quadratic  Surds. 

269.  A  quadratic  surd  is  the  indicated  root  of  an  imperfect 

square,  as  V2. 

270.  Theorem  1.  The  product  or  quotient  of  two  dis- 
similar quadratic  surds  will  be  a  quadratic  surd. 

Thus,  Vab  X  -Vatic  =  ab  ypj 

^s/abc  -*-  -\fa~b  =  Vc. 

Two  dissimilar  quadratic  surds  cannot  have  all  the  fac- 
tors under  the  radical  sign  alike.  Hence,  their  product  or 
quotient  will  contain  the  first  power  only  of  at  least  one 
factor,  and  will  therefore  be  a  surd. 

271.  Theorem  2.  The  sum  or  difference  of  two  dis- 
similar quadratic  surds  cannot  be  a  rational  number,  nor 
can  the  sum  or  difference  be  expressed  as  a  single  surd. 

For,  if  Va  ±  V#  could  equal  a  rational  number  c,  we 
should  have,  by  squaring, 

a±2Vab  +  b  =  c*; 
that  is,  ±  2  ^fab  =  c2  —  a  —  b. 

Now,  as  the  right  side  of  this  equation,  is  rational,  the 
left  side  would  be  rational ;  but,  by  §  270,  ~va~b  cannot  be 
rational.     Therefore,  Va  ±  V#  cannot  be  rational. 

In  like  manner  it  may  be  shown  that  Va  ±  V#  cannot 
be  expressed  as  a  single  surd  Vc. 


252  RADICALS. 

272.  Theorem  3.  A  quadratic  surd  cannot  equal  the 
sum  of  a  rational  number  and  a  surd. 

For,  if  Va  could  equal  c  -f  V#,  we  should  have,  by 
squaring, 

a  =  c*  +  2cVb  +  b, 
and,  by  transposing, 

2cVb  =  a-b-c2. 

That  is,  a  surd  equal  to  a  rational  number,  which  is 
impossible. 

273.  Theorem  4.  If  a  +  Vb  =  x  +  Vy,  then  a  will 
equal  x,  and  b  will  equal  y. 

For,  by  transposing,  V&  —  Vy  =  x  —  a ;  and  if  b  were 
not  equal  to  y,  the  difference  of  two  unequal  surds  would 
be  rational,  which  by  §  271  is  impossible. 

.-.  b  =  y,  and  a  =  x. 

In  like  manner,  if  a  —  "Vo  =  x  —  Vy,  a  will  equal  x, 
and  b  will  equal  y. 


274.    Theorem  5.     ijf  \a  +  V£  =   vS  4-  Vy,   **«» 

Y«  —  V#  =  Va!  —  Vy. 

Square  both  sides  of  the  .given  equation, 

a  +  Vb  =  x  +  2  Vxy  +  y. 

Therefore,  by  §  273,  a  =  x  +  y,  (1) 

and  V6  =  2VSy.  (2) 

Subtract  (2)  from  (1), 

a  —  Vft  =  x  —  2  Vxy  +  y, 

Extract  the  square  root  of  both  sides, 

V  a  —  V&  e=  Vz  —  Vy. 


RADICALS.  253 

275.    To  Extract  the  Square  Root  of  a  Binomial  Surd. 

1.  Extract  the  square  root  of  7  +  4  Vo. 

Let  Vx  +  y/y  =  V7  +  4V3.  (1) 

Then,  by  §  274,  Vi-Vy  =  V7-4V3.  (2) 

Multiply  (1)  by  (2),  x  -  y  =  V49  -  48. 

.\x  —  y  =  l. 
Square  (1),  then  §  273,        x  +  y  =  7. 

.-.  x  =  4,  and  y  =  3. 
.-.  Vx  +    Vy  =  Vi  -£  V3. 
.-.  V7  +  4V3  =  2  +  V3. 

A  root  may  be  found  by  inspection,  when  the  given  expression  can 
be  written  in  the  form  a  +  2  V6,  by  finding  two  numbers  that  have 
their  sum  equal  to  a  and  their  product  equal  to  6. 

2.  Find  by  inspection  the  square  root  of  75  —  12  V21. 

It  is  necessary  that  the  coefficient  of  the  surd  be  2 ;  therefore, 
75  —  12  V21  must  be  put  in  the  form  75  —  2V62  X  21 ; 

that  is,  75  — 2V756. 

Two  numbers  whose  sum  is  75  and  product  756  are  63  and  12. 

Then,  75  —  2  V756  =  63  —  2  V63  X  12  +  12 

=  (V63—  Vl2)2. 
That  is,  V63  —  Vl2  =  the  square  root  of  75  —  12  V2T ; 

or  3  V7  —  2  V3  =  the  square  root  of  75  —  12  V2I. 

3.  Extract  the  square  root  of  11  +  6  V2. 

11  +  6V2  =  ll  +  2Vl8. 
Two  numbers  whose  sum  is  11  and  product  18  are  9  and  2. 

'Then,       11  +  2  Vl8  =  9  +  2  V9  X  2  +  2 

=  (V9  +  V2)2. 
That  is,       V9  +  V2  =  the  square  root  of  11  +  6  V5; 
or  3  +  V2  =  the  square  root  of  11  +  6  V2. 


254  RADICALS. 

Exercise  105. 
Find  the  square  root  of : 

1.  7-4V3.  7.    16  +  5V7.  13.  94  +  42V5. 

2.  11  +  V72.  8.    75  +  12V2I.  14.  11-2V30. 

3.  7  +  2ViO.  9.    19  +  8V3.  15.  47-4V33. 

4.  18  +  8V5.  10.   8V6  +  20.  16.  29  +  6V22. 

5.  8+2VI5.  11.    28-16V3.  17.    83  +  12V35. 

* 

6.  15-4VH        12.    51+36V2.  18.    55- 12  V2l. 

Equations  Containing  Radicals. 

276.  An  equation  containing  a  single  radical  may  be 
solved  by  arranging  the  terms  so  as  to  have  the  radical 
alone  on  one  side,  and  then  raising  both  sides  to  a  power 
corresponding  to  the  order  of  the  radical. 


Solve  V#2-9  +  x  =  9. 


Transpose  x,  Vx2  —  9  =  9  —  x. 

Square,  x2  —  9  =  81  —  18  x  +  aft 

18s  =  90. 

.-.  x  =  5. 

277.    If  two   radicals  are  involved,  two   steps  may  be 
necessary. 

Solve  Va;  +  15  +  Vx  =  15. 

Square,  x  +  15  +  2  Vx2  +  15  x  +  x  =  225. 

Transpose,  2  Vx2  +  15  x  =  210  —  2  x. 

Divide  by  2,  Vx2+15x  =  105  -  x. 

Square,  x2  +  15  x  =  11026  —  210  x  +  aft 

225  x  =  11025. 
\  x  =  49. 


RADICALS. 


255 


Exercise  106. 
Solve: 

1.  2^x  +  5  =  V28.  8.    -^33  +  7  =  3. 

2.  3V4as~8  =  Vl3a!-a     9.   14  +  -^4  a;  -  40  =  10. 


3.  Vz  +  9  =  5Vo;  -3. 

4.  4  =  2Va--3. 

5.  5-V3^  =  4. 

6.  7  +  2^3^  =  5. 


7.    ^/2  x  -  3  =  -  3. 


10.  Vl0y-4  =  V7y  +  ll. 

11.  2V*-2  =  V32(a:-2)«. 

12.  V^4  +  x  =  |  +  V» 

13.  V32  +  a;  =  16  -  VS. 

14.  VS  —  Vaj  —  5  =  V5. 
15.    V»  +  20  -  -yJx  -  1  -  3  =  0. 


16.  Vx  +  15  -  7  =  7  -  Va;  -  13. 

17.  x  =  7  -  Va2  -  7. 


19. 


18.    Va-7  =  Vaj  +  1 

Va-  —  3       VS  +  1 

Va-  +  3  ""  Vz  -  2 


2i  i  +  q-*)* 

1  -  (1  -  x)* 


20.,- 


|-:->IR^»*^-s 


3)* 


23.  \/a  +  VS  +  ya  —  Va  =  VS. 

24.  Vox  —  1=4  +  £  Vox  —  -J. 

25.  3  Vie  —  3Va=  VS  —  Va  +  2Va. 

26.  V9  +  2a;-  V2x  = —j= 

V9  +  2a 


CHAPTER  XVm. 
IMAGINARY  EXPRESSIONS. 

278.  An  imaginary  expression  is  any  expression  which 
involves  the  indicated  even  root  of  a  negative  number. 

It  will  be  shown  hereafter  that  any  indicated  even  root 
of  a  negative  number  may  be  made  to  assume  a  form  which 
involves  only  an  indicated  square  root  of  a  negative  num- 
ber. In  considering  imaginary  expressions  we  accordingly 
need  consider  only  expressions  which  involve  the  indicated 
square  roots  of  negative  numbers. 

Imaginary  expressions  are  also  called  imaginary  numbers 
and  complex  numbers.  In  distinction  from  imaginary  num- 
bers all  other  numbers  are  called  real  numbers. 

279.  Imaginary  Square  Roots.  If  a  and  b  are  both  posi- 
tive, we  have 

Vab  =  Va  X  Vb. 
If  one  of  the  two  numbers  a  and  b  is  positive  and  the 
other  negative,  it  is  assumed  that  the  law  still  holds  true ; 
we  have,  accordingly : 

V^l  =  V4  (-  1)  =  VI  X 

V^5  =  V5  (- 1)  =  VE  x 

V^a  =  Va  (—  1)  =  Va  X 
and  so  on. 

It  appears,  then,  that  every  imaginary  square  root  can 
be  made  to  assume  the  form  a  V  —  1,  where  a  is  a  real 
number. 


IMAGINARY  EXPRESSIONS. 


257 


280.  The  symbol  V—  1  is  called  the  imaginary  unit,  and 
may  be  defined  as  an  expression  the  square  of  which  is  —  1. 

Hence,     V^l  X  V^  =  ( V^)2  =  -  1 ; 

V^a  X  V^~b  =  Va  X  V^l  X  Vb  X  V^l 

=  -\fab  X  (-  1) 
=  -  Vab. 

281.  It  will  be  useful  to  form  the  successive  powers  of 
the  imaginary  unit. 

(V^l) =  +  V=l; 

(V^)2 =-1; 

( V^i)8  =  ( V^l)2  V^I     =  (- 1)  V^i  =  -  V^i ; 

(V^l)«  =  (  V^l)2  (  V^T)2  =  (-  1)    (-  1)  =  +  1  ; 

(V^i)5  =  (V^T)4  V^i     =  (+  i)  V=i  =  +  vpij 

and  so  on.     If,  therefore,  n  is  zero  or  a  positive  integer, 

(V^~l)4n+1  =  +  ps/SI'j 

(V^T)4n+2=-l; 
(V^l)4n+3=- V^; 

(V^~i)4w+4=  +  i. 

Every  imaginary  expression  may  be  made  to  assume 
the  form  a  +  b  V—  1,  where  a  and  b  are  real  numbers,  and 
may  be  integers,  fractions,  or  surds. 

If  b  =  0,  the  expression  consists  of  only  the  real  part  a, 
and  is  therefore  real. 

If  a  =  0,  the  expression  consists  of  only  the  imaginary 
part  b  V—  1,  and  is  called  a  pure  imaginary. 


258  IMAGINARY  EXPRESSIONS. 

283.  The  form  a  -f  b  V—  1  is  the  typical  form  of  imaginary- 
expressions. 

Reduce  to  the  typical  form  6  +  V— 8. 
This  may  be  written  6  J-  Vs  X  V— 1,  or  6  +  2  V2  X  V^T ; 
here  a  =  6,  and  b  =  2  V2. 

284.  Two  expressions  of  the  form  a  •+■  b  V—  1,  a  —  b  V—  1, 
are  called  conjugate  imaginaries. 

To  find  the  sum  and  product  of  two  conjugate  imagi- 
naries : 


The  sum  is 


a  + W-l 
a  — &V-1 

2a 

a  -b   V^l 

a2  +  a&V^l 

The  product  is         a2  +  b2 

From  the  above  it  appears  that  the  sum  and  product  of 
two  conjugate  imaginaries  are  both  real. 

285.    Theorem  1.     An  imaginary  expression  cannot  be 
equal  to  a  real  number. 

For,  if  possible,  let 

a  ■+-  b  V—  1  =  c. 
Transpose  a,  #V—  1  =  c  —  a. 

Square,  —  b2  =  (c  —  a)'' 


J 


Since  J2  and  (c  —  a)2  are  both  positive,  we  have  a  nega- 
tive number  equal  to  a  positive  number,  which  is  impossible. 


IMAGINARY  EXPRESSIONS.  259 

286.  Theorem  2.  If  two  imaginary  expressions  are 
equal,  the  real  parts  are  equal  and  the  imaginary  parts  are 
equal. 

For,  let  a  +  b  V^T  =  c  -f  dV^T. 

Then,  (b  -  d)V^l  =  c  -  a; 

Square,  — (b  —  d)2  =  (c  —  a)2, 

which  is  impossible  unless  &  =  d  and  a  =  c. 

287.  Theorem  3.  Ifxandyarerealandx  +  yw  —  l  —  O^ 
then  x  =  0  awcZ  y  =  0. 

For,  y  V—  1  =  —  sc. 

Square,  —  y2  =  ic2. 

Transpose  —  y2,  sc2  +  y2  =  0, 

which  is  true  only  when  x  =  0  and  y  =  0. 

Operations  with  Imaginaries. 

1.  Add  5  +  7  VS-1  and  8  -  9  V^l. 

The  sum  is  5  +  8  +  7  V1^!  —  9V^1, 

13  —  2V:rT. 

2.  Multiply  3  +  2  V^l  by  5-  4V=:1. 
(3  +  2  V^T)  (5  -  4  vpl) 
=  15  —  12  V^~l  +  10  V^T  —  8  (—  1) 
=  23  — 2V^T. 

3.  Divide  14  +  5  V^l  by  2  -  3  V^l. 

14  +  5  V^T       (14  +  5  V^\)  (2  +  3  V^T) 
2-3V^T  ~~  (2-3V^T)(2  +  3V^l) 

_13_+52_V^T 
4-(-9) 

_13  +  52V^1 
13 

=  1  +  4V=:T, 


260 


IMAGINARY  EXPRESSIONS. 


Exekcise  107. 
Eeduce  to  the  form  b  V— 1 : 

i.  V^. 


2.    V-16. 


9.   V-625. 
10.   V-36. 


17.  V-  x18. 

18.  V^±. 


3.   V-25. 


11.   V-64. 


19.    V-  a4b~2. 


4.   V- 144.         12.   V-  729. 


20.   V-9x4. 


5.   V-169.  13.   V- 289.  21.   V-(2a?-3 


6.    V^ 


7.    V-81. 


8.   V-  256. 
Add: 


14.  7-1024. 

15.  V— a;8. 

16.  V-a9. 


22.  V-(«-2,y)2 

23.  V-  (x2  +  y2). 

24.  V-  (x2  -  y2). 


25.   V-  25  +  V-  49  -  V- 121. 


26.   V-  64  +  V^  -  V-  36. 


27.    V— a4  -f-  V-4a4  -f  V— 16  a4. 


28.  V^a2  +  V-  81  a2  -  V^ 

29.  a-bV^  +  a  +  bV^l. 

30.  2  +  3V=:l-2  +  3V^l. 

31.  a  +  bV^l  +  c-dV^. 

32.  SaV^  — (2a- ^V^. 
Multiply : 

33.  V^  by  V^. 

34.  -V^by  V^.  37.   V^2  by  V^^y5. 

35.  V-16  by  V^9.  38.   V^  by  V-16. 


36.   V-ic2  by  V^" 


IMAGINARY  EXPRESSIONS. 


261 


39.  V-  25  by  V-64.  41.  3V^3  by  2-^2. 

40.  V-  (a  +  b)  by  V-  (a  -  b).  42.   -  5V^2  by  2  VhS. 

43.  V^  +  V^  by  V^  -  V^5. 

1  +  V-^3  _j  1  -  V^3 

44.  jt by  x g 

45.  a  V—  a  +  b  V—  #  by  a  V—  a  —  J  V—  6. 

46.  2V^2  +  3V^  by  S-^l  —  2V^~5. 

47.  -y/3  +  2^3  by  V3-2V^3. 

48.  m  —  3  V—  6  by  w  +  4  V—  c. 

Perform  the  divisions  indicated : 


49. 

a 

50. 

b 

V-fc2 

51. 

G 

V=4 

52. 

V=9 

V-81 

53. 
54. 

V=6 

V—  ax 

55. 


56. 


57. 


58. 


59. 


60. 


V-a? 


V- 


V-lOx8 

V— 5cc 


8V-a;2 

2Vx 


61. 

1 

3- V^2 

62. 

2  + V^2 

i'-V=I 

63. 

a+x V— 1 

a  —  ccV—  1 

64. 

VB  + V^6 

V6- V^~8 

CK 

2a  +  36V^l 

66. 


2a-3&V^T 
4a-£&V^l 


CHAPTER  XIX. 
QUADRATIC  EQUATIONS. 

288.  We  have  already  considered  equations  of  the  first 
degree  in  one  or  more  unknowns.  We  pass  now  to  the 
treatment  of  equations  containing  one  or  more  unknowns 
to  a  degree  not  exceeding  the  second.  An  equation  which 
contains  the  square  of  the  unknown,  but  no  higher  power, 
is  called  a  quadratic  equation. 

289.  A  quadratic  equation  which  involves  but  one  un- 
known number  can  contain  only  : 

1.  Terms  involving  the  square  of  the  unknown  number. 

2.  Terms  involving  the  first  power  of  the  unknown 
number. 

3.  Terms  which  do  not  involve  the  unknown  number. 

Collecting  similar  terms,  every  quadratic  equation  can  be 
made  to  assume  the  form 

ax2  -f-  bx  -f-  c  =  0, 

where  a,  b,  and  c  are  known  numbers,  and  x  the  unknown 
number. 

If  a,  b,  c  are  numbers  expressed  by  figures,  the  equation 
is  a  numerical  quadratic.     If  a,  b,  c  are  numbers  represented   i 
wholly  or  in  part  by  letters,  the  equation  is  a  literal  quadratic. 


In  the  equation  ax2  -f-  bx  -+■  c  =  0,  a,  b,  and  c  are  called 
the  coefficients  of  the  equation.  The  third  term  c  is  called 
the  constant  term. 


QUADRATIC  EQUATIONS.  263 

290.  If  the  first  power  of  x  is  wanting,  the  equation  is  a 
pure  quadratic ;  in  this  case  b  =  0. 

If  the  first  power  of  x  is  present,  the  equation  is  an 
affected  or  complete  quadratic. 

Pure  Quadratic  Equations. 


1.    Solve  the  equation 

5x* 

-48 

=  2x*. 

Collect  the  terms, 

3x2  = 

48. 

Divide  by  3, 

X*  = 

16. 

Extract  the  square  root, 

X  = 

±4. 

It  will  be  observed  that  there  are  two  roots,  and  that  these  are 
numerically  equal,  but  of  opposite  signs.  There  can  be  only  two 
roots,  since  any  number  has  only  two  square  roots. 

It  may  seem  as  though  we  ought  to  write  the  sign  ±  before  the  x 
as  well  as  before  the  4.  If  we  do  this,  we  have  +  cc  =  +  4,  —  x  =  —  4, 
+  x  =  —  4,  —  x  =  +  4. 

From  the  first  and  second  equations,  x  =  4 ;  from  the  third  and 
fourth,  x  =  —  4 ;  these  values  of  x  are  both  given  by  the  equation 
x  =  ±  4.  Hence  it  is  unnecessary  to  write  the  ±  sign  on  both  sides  of 
the  reduced  equation. 

2.  Solve  the  equation  3  x2  —  15  =  0. 

Transpose,  3  x2  =  15. 

Divide  by  3,  a;2  =  5. 

Extract  the  square  root,  x  =  ±  V5. 

The  roots  cannot  be  found  exactly,  since  the  square  root  of  5  can- 
not be  found  exactly ;  it  can,  however,  be  determined  approximately 
any  required  degree  of  accuracy ;  for  example,  the  roots  lie  between 
.23606  and  2.23607  ;  and  between  —  2.23606  and  —  2.23607. 

3.  Solve  the  equation  3  x2  +  15  =  0. 

Transpose, 

Divide  by  3, 

Extract  the  square  root, 


264  QUADRATIC  EQUATIONS. 

There  is  no  square  root  of  a  negative  number,  since  the  square  of 
any  number,  positive  or  negative,  is  necessarily  positive. 

The  square  root  of  —  5  differs  from  the  square  root  of  +  5  in  that 
the  latter  can  be  found  as  accurately  as  we  please,  while  the  former 
cannot  be  found  at  all. 

291.  A  root  that  can  be  found  exactly  is  called  an  exact 
or  rational  root.  Such  roots  are  either  whole  numbers  or 
fractions. 

A  root  that  is  indicated  but  can  be  found  only  approxi- 
mately is  called  a  surd.  Such  roots  involve  the  roots  of 
imperfect  powers. 

Eational  and  surd  roots  are  together  called  real  roots. 

A  root  that  is  indicated  but  cannot  be  found,  either 
exactly  or  approximately,  is  called  an  imaginary  root.  Such 
roots  involve  the  even  roots  of  negative  numbers. 

Exercise  108. 
Solve : 

1.    3x2-2  =  x2  +  6.  n     3  -  x2  ,  x2  +  5      0 

9.    — —         I 7, —  —  o. 


2.    5a2-r-l0  =  6z2  +  l. 


11       '        6 


3.  7 x2- 50  =  Ax2 +  25.  5x2  +  3      17  -  x2 

1  11  84 

4.  6x2  —  -  =  4ic2  +  tt' 

6  9  3  17 

„     x2  +  l      ,A  U:    4z2      6z2-3* 

5.  — - —  =  10. 

5 


5 


6.    3g;2~8  =  4.  12'    3z2      5x2      15 


10 
x2-9   _  x2  +  l  i3. 


1 


7-    — 4—  =  — 5 '    x-1      x  +  1      4 

Q     2^2-4  ,  o:2  +  4  15'-           7       _n 

8. —  =  8.  14. h  7: r—  =  Z. 

7                 5  8  —  x      2  —  3x 


QUADRATIC  EQUATIONS.  265 

15.  3ft24-llft  =  10ft4-84-ft24-a. 

16.  (ft  +  4)  (x  +  5)  =  3  (x  +  1)  (x  +  2)  -  4. 

17.  3  (a  -  2)  (ft  +  3)  =  (ft  +  1)  (ft  +  2)  -h  a;2  +  5. 

18.  (2  ft  +  1)  (Sx  -  2)  +  (1  -  x)  (3  +  4ft)  =  3  ft2  -  15. 
ft2 +  9      2x*-5  ,  3x*  +  10      ,. 

3  ft2 -5  ,   2a;2 +  4      ft2  -  3      M 
20.   —^-  +  —9 J—* 

10ft2  +  7      12a;2  +  2  _  5a:2  -  9 
r  18  lift2- 8  ~        9 

nn    x  —  1   ,  ft  + 1      5  „      a  ,  ft      9  a2  —  ft2 

22.  — — r  + r  =  o-  26.    -  +  -  = 

ft  + 1      ft  — 12  ft      a  aa; 

23.  aft2  +  &  =  c.  oiy    fl  +  a   ,  ft  — a_5 

ft  —  a      ft  +  a      J 

24.  aft2  +  ^  =  ^2  +  a. 

2ft       5*  +  2ft_ 

25.  ft2  +  2to  +  c  =  &(2ft  +  l).  ft-£>  3ft 

29.  2|(ft  +  a)  (ft  +  5)  +  (*  -  a)  (ft  -  5)j  =  a2  4-  4  J2. 

30.  2  ^  (ft  -  a)  (ft  4-  &)  +  (ft  4-  a)  (x  -  b)  \  =  9  a2  4-  2  ab  +  b\ 

Affected  Quadratic  Equations. 

292.  Since  (x  ±  b)2  =  ft2  =fc  2  Jft  4-  J2,  it  is  evident  that 
the  expression  ft2  ±2  foe  lacks  only  the  third  term,  b2,  of 
being  a  perfect  square. 

This  third  term  is  the  square  of  half  the  coefficient  of  x. 

Every  affected  quadratic  may  be  made  to  assume  the 
form  a;2  ±  2  bx  =  c,  by  dividing  the  equation  through  by  the 
coefficient  of  ft2. 


I 


266  QUADRATIC  EQUATIONS. 

293.    To  solve  such  an  equation : 

The  first  step  is  to  add  to  both  members  the  square  of 
half  the  coefficient  ofx.    This  is  called  completing  the  square. 

The  second  step  is  to  extract  the  square  root  of  each  mem- 
ber of  the  resulting  equation. 

The  third  step  is  to  reduce  the  two  resulting  simple 
equations. 

1.    Solve  the  equation  x2  —  8  x  =  20. 

Complete  the  square,  x2  —  8  x  +  16  —  36. 

.  Extract  the  square  root,  x  —  4  =  ±  6. 

Reduce,  x  =  4  +  6  =  10, 

or  x  =  4  —  6  =  —  2. 

The  roots  are  10  and  —  2. 

Verify  by  putting  these  numbers  for  x  in  the  given  equation. 


x  =  10, 

102  -  8  (10)  =  20, 

100  -  80  =  20. 


x  =  -2, 
(-  2)2  -  8  (-  2)  =  20, 
4  +  16  =  20. 


ooi      xi-  ,.-       x  -hi       4cc  —  3 

2.    Solve  the  equation = — —  • 

*  x  —  1        a;  -f-  9 

Free  from  fractions,  (x  +  1)  (x  +  9)  =  (a;  —  1)  (4x  —  3). 
Simplify,  —  3  x2  +  17  x  =  —  6. 

Divide  by— 3,  x2  —  ±£x  =  2. 

Half  the  coefficient  of  x  is  £  of  —  J^- =  —  V>  and  the  square  of  —  *j 
is  ^9-.     Add  the  square  of  —  ^  to  both  sides,  and  we  have 


X         3    fUJ       "^  36        36 


Extract  the  root, 
Transpose  —  ty 


17 
6 

■?.*»■ 

X 

17  ±19 
6 

• 

.'.  X 

=  6,  or  — 

1 

QUADRATIC  EQUATIONS.  267 


EXERCISI 

s  109 

Solve : 

1.   cc2  +  2ic  =  8. 

7. 

2a;2 -fa;  =  15. 

2.   x2-6x  =  7. 

8. 

5a;2  +  3a;  =  2. 

3.   a;2 -4a;  =  12. 

9. 

a;2  +  \x  =  40. 

4.   jc2-f-4a;  =  5. 

10. 

3a;2 -4a;  =  4. 

5.   a;2  +  5a;  =  14. 

11. 

6a;2  +  a;  =  l. 

6.   a;2 -3a;  =  28. 

12. 

6  a;2  -x  =  2. 

13.    12a;2 -11a;- 

4-2 

=  0. 

14.    15x2-2x- 

1  = 

:0. 

{x  +  1){x  +  2) 

(x- 

■l)(--2)      3 

5 

2 

(2  a; -3)  a; 

(*  +  4)  ( 

*-l)      . 

1C'            4 

6 

17. 

3a  +  5      2a;-5      0 
a; +  4    '    a? -2 

22. 

a;  +  2      4-a;      7 
a;-l         2a;        3 

18. 

a;  —  6       a;  4-  5        . 
a; -2  f  2a;  +  1 

23. 

a;  +  3      5a;  +  8 
a;  — 2        a;  +  4 

19. 

4-3a;      l  +  2a;      9 
2  +  x         1-x       2 

24. 

2a;-l            3 
3           2a;-l~ 

20. 

x         x  +  1      13 
a;  + 1          x           6 

25. 

x+1      x+2      13 
a;  +  2      a;  +  l~  6 

21. 

5  a;2  —  4  a;  =  1. 

26. 

7a;2-8a;  =  -l. 

268  QUADRATIC  EQUATIONS. 

294.  If  the  coefficient  of  x2  is  4,  9,  16,  or  any  other 
perfect  square,  we  may  complete  the  square  by  adding  to 
each  side  the  square  of  the  quotient  obtained  from  dividing 
the  second  term  by  twice  the  square  root  of  the  first  term. 

Solve  4«2  -  23  x  =  -  30. 

23 
The  square  root  of  4a;2  is  2x,  and  23 x  divided  by  twice  2x  is  —  • 

23 
Add  the  square  of  —  to  both  sides. 

Then,  ^-23*+(fy  =  f-30  =  f 

23  7 

Extract  the  root,  2x  — —=  ±  -' 

4  4 

_  23  ±7      30   ftr16 

Transpose,  2  x  =  — - —  =  — '  or  "T" 

.-.  x  =  3f,  or  2. 

If  the  coefficient  of  x2  is  not  a  perfect  square,  we  may 
multiply  the  equation  by  a  number  that  will  make  the 
coefficient  of  x2  a  perfect  square. 

Solve  -3x2  +  5x  =  -2. 

Since  the  even  root  of  a  negative  number  is  impossible,  it  is  neces- 
sary to  change  the  sign  of  each  term.     The  resulting  equation  is 

3x2  —  bx  =  2. 
Multiply  by  3,  9  x2  —  15  x  =  6. 

Complete  the  square,     9x2  —  15jc  +  —  =  —  • 


Extract  the  square  root, 

»H=±l 

Reduce, 

*-*p- 

3x  =  6,  or  —  1. 

.-.  x  =  2,  or  -  -  • 

QUADRATIC  EQUATIONS.  269 

Exercise  110. 


Solve : 


1.  3a2-2a  =  8.  2,2x      OK 

14.   3  x2  +  —  =  25. 

2.  5a2 -6a  =  27. 

3.  2a2  +  3a  =  5.  15'   *"--f  =  3*  +  l. 

4.  2a2-5a  =  7.  16.   ^  -  |  =  2  (^  -  2). 

5.  3a2  +  7a  =  6. 

6.  5a2 -7a  =  24. 

7.  8a2 -h  3a  =  26. 

8.  7  a2  +  5a  =  150. 

9.  6a2  +  5a  =  14. 

10.  7a2  -  2a  =  £. 

11.  8a2  +  7a  =  51. 

12.  7a2 -20a  =  75. 

13.  11  a2- 10  a  =  24.  3        2a;       3 

23.  (x  +  2)  (2  a  +  1)  +  (x  -  1)  (3a  +  2)  =  57. 

24.  3a  (2a  +  5)  -  (a  +  3)  (3a  -  1)  =  1. 

25     (2a  +  5)(a-3)      a(3a  +  4)_5> 
3  5 

26.   £  (5a2  -8a-  6)-  i  (a2-  3)  =  2a  +  1. 

L   •+*,.*        29.  -i-^^-i-. 

a;  +3      a:  a—  1      as—  2      a;—  4 

28    _J___3 4  g+,2     a  +i3 


17. 

2a2  ,  3a      +M 

18. 

3*      „   2      1 

T-2a=i6 

19. 

|a2  +  |a  =  y 

20. 

2a-3  =  -- 

X 

21. 

7  a       5        20 
5        3a~  3  * 

270  QUADRATIC  EQUATIONS. 


Another  Method  of  Completing  the  Square. 

295.  If  a  complete  quadratic  is  multiplied  by  four  times 
the  coefficient  of  x2,  fractions  will  be  avoided. 

So\ve3x2-5x  =  2. 

Multiply  by  12,  36  x2  —  60  x  =  24. 

Complete  the  square,         36  x2  —  60  x  +  25  =  49. 
Extract  the  square  root,  6  x  —  5  =  ±  7. 

Reduce,  6x  =  5±7. 

6  x  =  12,  or  —  2. 

.-.  x  =  2,  or  —  -  • 

The  number  added  to  complete  the  square  by  this  last  method  is 
the  square  of  the  coefficient  of  x  in  the  original  equation  3  x2  —  5  x  =  2. 

296.  If  tbe  coefficient  of  x  is  an  even  number,  we  may 
multiply  by  the  coefficient  of  x2,  and  add  to  each  member 
the  square  of  half  the  coefficient  of  x  in  the  given  equation. 

Solve  3  z2  + 4  a  =  20. 

Multiply  by  the  coefficient  of  x2  and  add  to  each  side  the  square  oi 
half  the  coefficient  of  x, 

9x2  +  (    )  +  4  =  64. 
Extract  the  square  root,  3  x  +  2  —  ±  8. 

Reduce,  3  x  =  —  2  ±  8. 

3  x  =  6,  or  —  10. 
.-.  x  =  2,  or  —  3f 

Note.  If  a  trinomial  is  a  perfect  square,  its  root  is  found  by  taking 
the  square  root  of  the  first  and  third  terms  and  connecting  these  roots 
by  the  sign  of  the  middle  term.  It  is  not  necessary,  therefore,  in  com- 
pleting the  square,  to  write  the  middle  term,  but  its  place  may  be 
indicated  by  a  parenthesis,  as  in  this  example. 

Verify  by  putting  the  values  of  x  in  the  given  equation. 

x  =  2.  x  =  —  3i. 

3  (2)2  +  4  (2)  =  20.  3  (-  8i)2  +  4  (-  3*)  =  20. 

12  +  8  =  20.  33*  -  13*  =  20. 


QUADRATIC  EQUATIONS.  271 

Exercise  111. 


Solve 


2x  ,    1       „  x  +  1      2x 


L   **       3   +12~°-  4*   ^+4_  a^  +  6 

a;2      a;  a;  a  +  3  1 

2.    ---  =  2(*  +  2).  5.    ______=__, 

3£     _4_      JS  2  3  2 

*     4+33:       6*  «-l~.z-2      *-4* 

3a?  5        3a2  23 


2(a?  +  l)      8      x2-l      4(jc  — 1) 
g    11-303  |  2(7-4g)_L 


..^+* 


a32-4   '   a;  +2      5(0  —  2) 

10. -  H — —  =  5. 

2  x  —  3        a;  -f-  1 


2a;  +  3  1-x     =  7-3a? 

*   2(20-1)      2(a;^-l)~4-3a;', 

--fc    2».— 1  3  a;-2  ,   , 

12'   — ^73^  =  ^8  +  5. 

3a:  +  2        7  -  ar  _  7s-l 
2#-l'r2a;-f-l      4»a-l"1" 

x  —5  ;  0—8^     SO        1 

'   a;+3a5-3_a;2-92' 

M2gfl     4a; +  1  45 

7-a:         7  +  a;        49 -a2"*"1' 


272  QUADRATIC  EQUATIONS. 


Solution  by  Resolving  into  Factors. 

297.  A  quadratic  which  has  been  reduced  to  its  simplest 
form,  and  has  all  its  terms  written  on  one  side,  may  often 
have  that  side  resolved  by  inspection  into  factors,  and  the 
roots  found  by  putting  each  factor  equal  to  zero. 

1.  Solve  x2  +  7  x-  60  =  0. 

Since  x2  +  7  x  —  60  =  (x  +  12)  (x  —  5),  (§  130) 

the  equation  x2  +  7  x  —  60  =  0 

may  be  written  (x  +  12)  (x  —  5)  =  0. 

If  either  of  the  factors  x  +  12  or  x  —  5  is  0,  the  product  of  the  two 
factors  is  0,  and  the  equation  is  satisfied. 

Hence,  x  +  12  =  0,  or  x  —  5  =  0. 

.-.  x  =  —  12,  or    x  =  6. 

2.  Solve  xs  -  x2  -  6x  =  0. 

The  equation  x3  —  x2  —  6  x  =  0 

may  be  written  x  (x2  —  x  —  6)  =  0, 

or  x(x  -  3)  (x  +  2)  =  0,                           (§  130) 

and  is  satisfied  if  x  =  0,  3,  or  —  2. 

Hence,  the  equation  has  three  roots,  0,  3,  —  2. 

3.  Solve  xs  -  2 x2  -  Vlx  +  12  =  0. 

By  the  Factor  Theorem  (§  135),  we  find  that  1  put  in  place  of  x 
satisfies  the  equation,  and  is  therefore  a  root  of  the  equation. 
Divide  by  x  —  1,  and  resolve  the  quotient  into  its  factors. 
We  have  (x  -  1)  (x  -  4)  (x  +  3)  =  0. 

Hence,  the  roots  of  the  equation  are  1,  4,  —  3. 

4.  Solves*  +  3 x  -10  =  0. 

If  we  add  ->  the  square  of  half  the  coefficient  of  x,  to  the  first  two 

9 
terms,  we  have  a  perfect  trinomial  square.     Add  and  subtract  -> 


QUADBATIC  EQUATIONS.  273 


x2  +  3x  +  |-|-10  =  0, 
3\2      49 


x  +  -j  —  —  =  0. 
We  now  have  the  difference  of  two  squares,  and  the  factors  are 

that  is,  (x  +  5)  (x  —  2)  =  0. 


.-.  x  =  —  5,  or  2. 
5.    Solve  3z2-  2#  -2-0. 

Divide  by  3,  x2  —  \x  —  §  =  0. 


Add  and  subtract  the  square  of  half  the  coefficient  of  x, 

*-!»+(f)'-l-N 

/        1\2      7 
that  is,  ^x  —  -J  —  -  =  0. 

7  /l  -i 

The  square  root  ofq  =  A/-X7  =  i"^* 

Hence,        (x  -|  +  |Vf)  (x  -  |-|V?)  =0. 

Therefore,  x  =  -  —  -  V7,  or  -  +  r  V7. 

6.    Solve  a2  -  a;  +  1  =  0. 

x2-x  +  7-4  +  l=0. 
4      4 


ference  of  tw( 

(HH-fy 


In  order  to  make  this  the  difference  of  two  squares,  write  it 
1\2      /      3" 


3  R  1      

The  square  root  of  —  -  =  a/-  X  (—  3)  =  ^—3- 

Hence,         (x  -\  +  \V=j)  (x  -\  -|V=s)  =0. 

Therefore,        x  =  | -  1  V=8,  or  |  +  1  V=3. 


274  QUADEATIC  EQUATIONS. 

Exercise  112. 
Resolve  into  factors,  and  find  the  values  of  x : 

1.  x2-  5x  +  4  =  0.  5.   xs  -f  x2-6x  =  0. 

2.  6x2-5x-6  =  0.  6.   a8 -8  =  0. 

3.  2x2-  x  -3  =  0.  7.   a8 +  8  =  0. 

4.  10x2  +  a;-3  =  0.  8.    x*  -  16  =  0. 

9.  (a?  -  1)  (x  -  3)  (a2  +  5x  +  6)  =  0. 

10.  (2x  -l)(x-  2)  (3ar>  -  5x  -  2)  =  0. 

11.  (x2  +  aj  -  2)  (2  a2  +  3  x  -  5)  =  0. 

12.  z8  +  ;c2-4(z  +  l)  =  0. 

13.  3z8  +  2x2-(3a;  +  2)  =  0. 

14.  x8  -  27  -  13  (a*  -  3)  =  0. 

15.  a8  +  8  +  3(;z2-4)  =  0.      17.   2x*-  2x'-  (x*-  1)-  0. 

16.  x(x2-l)-6(x-l)  =  0.  18.   *8-3a:-2  =  0. 

19.    2cc8  +  2a;2  +  (x2-5a;-  6)  =  0. 

298.  Any  quadratic  trinomial  of  the  form  ax2  +  bx  +  o  can 
be  resolved  into  two  factors  by  writing  it  as  the  difference 
of  two  squares.     Thus, 

Sx2  +  7 x  -  6  =  3  (x2  +  %  x  -  2) 

=  3[(z  +  £)2-W] 

=  3(a  +  S  +  V)(*  +  £-J5L> 
=  3  (x  +  3)  (a  -  3) 
=  (aj -(-3)  (3  a; -2). 


QUADRATIC  EQUATIONS.  275 

Exercise  113. 
Resolve  into  factors  : 

1.  8x2-26x  +  21.  5.    5«2-15z  +  ll. 

2.  6x2-x-7.  6.    3x2  +  5x-2. 

3.  2x2  +  7x  +  6.  7.    3aj2-cc-l. 

4.  5a2  +  26a +  24,  8.    x2-Sx-5. 

Literal  Equations. 

1.  Solve  the  equation  adx  —  acx2  =  hex  —  bd. 

Transpose  hex  and  change  the  signs, 

acx2  +  bcx  —  adx  —  bd. 
Express  the  left  member  in  two  terms, 

acx2  +  (be  —  ad)x  =  bd. 
Multiply  by  4  times  the  coefficient  of  x2, 

4  a2c2x2  +  4  ac  (be  —  ad)x  =  4  abed. 
Complete  the  square, 

4  a2c2x2  +  (    )  +  (bc-ad)2  =  b2c2  +  2  abed  +  a2<P. 
Extract  the  root,         2  acx  +  (6c  —  ad)  =  ±  (be  +  ad). 
Reduce,  2  acx  =  —  (6c  —  ad)  ±  (6c  +  ad) 

=  2  ad,  or  —  2  be. 

d  6 

.-.  x  =  -i  or 

c  a 

2.  Solve  the  equation  px2  —  px-\-  qx2  +  qx  —  -*■■ — • 

(p  +  g)x2-(p-g)x=^j^- 

4  (p  +  g)2x2  —  4  (p2  —  g2)  x  =  4pg. 

4(p  +  g)2x2-(    )  +  (p-g)2=p2  +  2pg  +  g2. 

2  (P  +  g)  x  —  (p  -  q)  =  ±  (p  +  g). 

2(p  +  g)x  =  (p  -  q)  ±  (p  +  q). 

P  Q 

.-.  x  =      ,   -  or *■—  • 

p  +  g  p+q 

Note.  The  left-hand  member  of  the  equation  when  simplified 
must  be  expressed  in  two  terms,  simple  or  compound,  one  term  con- 
taining x2  and  the  other  term  containing  x. 


276 


QUADRATIC  EQUATIONS. 


Exercise  114. 


Solve : 

1.  x2  +  2ax  =  3a2. 

2.  x2  -  4  ax  =  12  a2. 

3.  x2  +  8bx  =  9b2. 

4.  x2  +  3bx  =  10b2. 

6.  x2  +  5ax  =  Ua2. 

6.      3X2  +  4:CX   =  4:C2. 

7.  5ax-2x2  =  2a2. 

8.  6a2-a;c-a2  =  0. 

x2  -{-  ax  +  a9 


9.    2  a2x2  +  ax  -  1  =  0. 
10.    12  b2x2-  5bx  =  3. 


2x? 


ax 


11. 

—  +  T  =  lla( 

12. 

3x2       x         3 

4    +2a~2a2 

13. 

2      •        3 
a      4r 

3  ax2  ,  2x      13 
14'   -4-  +  T  =  37 


15. 

16.    2cc2-a  +  a 
x  (a  —  x)    .  x 


xz 


3 

■■2  a2. 


ax. 


17. 


18. 


+  3  =  5«. 


a  4"  X 

x(3x  —  a)  _  a 


19.    x2-\ 
20. 


4#  +  a 

m2  —  ?is 


6b2 


mn 
5x 


12 

cc  =  l. 
1 


4 

22.    a2  +  (a  -  b)  x  =  ab. 
2a  +  x      a-2x      8 
2a-a;      a  +  2a;~~3' 
2#-3a  .  3<r4-2a      10 
7' 


23 


24. 


bab      az 


0  —  a      x  4-  a 


25. 


26. 


27. 


sc  +  4a 
a  +  26 
36-a 

<7 


4ic  —  a 

a  +  5 


«  +  2a 


b  -{•  x      a  -\-b 
~~      x  +  b 


a  -\-  b  -\-  x 

1 
a  +  b  ■+■  x 


2. 


a      6      sc 


28.  9ic2-3(a  +  2&)*  +  2a6  =  0. 

29.  (2  a  +  l)a;2  +  3  a2x  +  a*  -  a2  =  0. 

30.  (1  -  a2)  x2  -  2  (1  4-  a2)  x  +  1  —  a2  =  0. 

31.  (a  +  b)2x2  -  (a2-b2)x  =  ab. 

32.  (a  4-  &)  a2  -  (2  a  +  £)  a  +  a  =  0. 


33. 

34. 
35. 
36. 
37. 
38. 

39. 

47. 
48. 
49. 
50. 

51. 
52. 
53. 
54. 
55. 
56. 


x      x  +  b      a 
1  1 


QUADRATIC  EQUATIONS. 
1 


277 


a  +  b 

_3  +  x2 
a  —  x      a  +  x      a2  —  x2 

x2-\-2ab(a2-\-b2)       0 
a2  +  b2  -**- 

(2x-ay     _ 
2x-a  +  2b~    ' 

(a-iyx*  +  2(3a-l)x 
4a -1 


x 


x  —  a 

ex  =  ax2  +  bx2 


2c 
b      x  —  c 

ac 


40.  ^  =  a  -j —  • 

x  —  3  x  4-  3 

41.  mar  —  1  =  — » L 

mn 

42.  (odi  -  5)  (bx  -  a)  =  c2. 
ax  -\-b      mx  +  w 

4o. 
44. 
45. 
46. 


bx  + 

a       7l£C  ■+-  m 

m 

1       ^ 

m  + 

cc      m  —  x 

& 

(••- 

-  5s2)  (x2  +  1) 

a24-52 

«2- 

4mraas 

a  +  5 

|  (a2  4-  a2  4-  a5)  =  J*  (20a  +  45). 
jc2  —  2  mcc  =  (n  —  p  -\-  m)  (n  —  p  —  m). 
a2  —  (m  +  w)  a  =  I  (p  +  a  +  m  +  w)  (p  +  q  —  m  —  n). 
mnx2  —  (m  4-  n)  (mn  4- 1)  x  4-  (m  4-  w)2  =  0. 
x2  m2  —  4:  a2      x 


3m  — 2a      4a  —  6m      2 

R      ,   (a  +  5)2      K/        .v,  25 aft 

6  x  4-  * *-  =  5  (a  —  5)  4- 

x 


2b-x-2a   x  45 
a  —  25  —  x 


6x 
x  —  4a 


ax  —  bx       ab  —  52 
55  —  x     .  2a  — a;  — 195 


a2  —  4  52         ace  4-  2  5x 


2bx 


a-25 
a;  4-25 


cc  +  13a4-35 
5a  —  35  ^-  sc 

a  +  35 
Sa2-12ab      9b2-±a2      (2a  +  35)  (x  -  35) 


35 


a4-35 


278  QUADRATIC  EQUATIONS. 

Solution  by  a  Formula. 

299.   Every  affected  quadratic  can  be  reduced  to  the  form 

ax2  +  bx  +  c  =  0. 

Solve  ax2  +  bx  +  c  =  0. 

Transpose  c,  ax2  +  bx  =  —  c. 

Multiply  the  equation  by  4  a  and  add  the  square  of  6, 

4a2a:2+(    )  +  62  =  62-4ac. 
Extract  the  root,  2ax  +  b=  ±  V62  —  4  ac# 


—  &  ±  V&2  —  4  ac 
•*-X  = 2^ ■• 

By  this  formula,  the  values  of  x  in  an  equation  of  the 
form  ax2  +  bx  4-  c  =  0  may  be  written  at  once. 

Solve  Sx2-  5a +  2  =  0. 

Here  a  =  3,  6  =  —  5,  c  =  2. 

Putting  these  values  for  the  letters  in  the  above  formula,  we  have 


5  +  V25  —  24        5— V25  — 24 
X  = j .or - 

=  I,  or  f 
=  1,  or  f. 

Exercise  115. 
Solve  by  the  above  formula : 

1.  2aja  +  3aj  =  14.  7.  5x2  -Ix  =  -2. 

2.  Sx2-5x  =  12.  8.  4x2-9x  =  28. 

3.  <c2-7a  =  18.  9.  5x2+Tx  =  12. 

4.  5z2-a;  =  42.  10.  II*2  -  9x  =  - -g- 

5.  6z2-7a;  =  10.  11.  7*2-f-5x  =  38. 

6.  3x2-lla;  =  -6.         12.  5x2-7x  =  & 


QUADRATIC  EQUATIONS.  279 

Equations  Involving  Two  or  More  Radicals. 


JOO.    Solve  Vx+i  +  V2x  +  6  =  V7  x  -f- 14. 


Square,  x  +  4  +  2V(x  +  4)(2x  +  6)  +  2  x  +  0  =  7  x  +  14. 
Simplify,  V(x  +  4)(2x  +  t>)  =  2x-f2. 

Square,  (x  +  4)  (2  x  +  C)  =  (2  x  +  2)2. 

Solve,  x  =  5,  or  —  2. 

Of  these  two  values,  only  5  will  satisfy  the  original  equation. 

Squaring  both  members  of  the  original  equation  is  equivalent  to 
transposing  V7x  +  14  to  the  left  member,  and  then  multiplying  by 
the  rationalizing  factor  Vx  +  4  +'V2x  +  G  +  V7x  +  14,  so  that 


(Vx  +  4  +  V2x  +  G  —  V7x+14)(Vx  +  4  +  V2x  +  6  +  V~7x  +  14)=0, 


and  this  reduces  to  V(x  +  4)(2x  +  0)  —  (2  x  +  2)  =  0. 

Transposing  and  squaring  again  is  equivalent  to  multiplying  by 


(Vx  +  4  —  V2  x  +  6  +  V7x  +  14)(Vx  +  4  —  V2x  +  C  —  V7x+  14). 
Reducing,  x2  —  3  x  —  10  =  0. 

Therefore,  the  equation  x2  —  3  x  —  10  =  0  is  really  obtained  from 


(Vx  +  4  -f  V2x  +  6  -  vTx  +  14) 


X  (Vx  +  4  +  V2x  +  6  +  V7 x  +  14) 


X  (Vx  +  4  -  V2x  +  6  —  V7x+  14) 

X  (Vx  +  4  —  V2x  +  6  +  V7 x  +  14)  =  0. 

This  last  equation  is  satisfied  by  any  value  that  will  satisfy  any  one 
)f  the  four  factors  of  its  left  member.  The  first  factor  is  satisfied  by 
6,  and  the  last  factor  by  —  2,  while  no  values  can  be  found  to  satisfy 
le  second  or  third  factor. 

As  5  is  the  only  value  of  x  that  will  satisfy  the  original  equation, 

other  values  must  be  rejected. 


280  QUADRATIC  EQUATIONS. 

Exercise  116. 
Solve  : 

1.  V9  x  +  40  -  2  Va;  +  7  =  Val 

2.  Va  +  a;  +  Va  —  x  =  V#. 


o     3x+W4:X-x2      0 

3.  =  —  J. 
3  ic  —  V4  a;  —  x2 

4.  Va;  -  3  -  Va;  -  14  =  V4a;  -  155. 

5.  VaT+4  —  Va:  =  VaT+|. 

3Vc-4      15  +  3Vs 
2+Vz         40+Vc 

7.   V14  a;  +  9  +  2  Va7+1  t  V3  a;  +  1  =  0. 


8.   V5a;  +  1  -  2  —  Vaj  +  1  =  0. 


9.   Va;  -  2  +  Va;  +  3  —  V4a;  +  1  =  0. 


10.   V7  -  a;  +  V3a;  +  10  +  Va;  +  3  =  0. 


11.   3Vc8  +  17  +  Va;8  +  1  +  2V5a;8  +  41  =  0 


12.    2x—W2x-l=x  +  2. 

"2-V2a^  =  0. 
1 


12. 


-  V<ca-8 


QUADRATIC  EQUATIONS.  281 


Equations  in  the  Quadratic  Form. 

301.  An  equation  is  in  the  quadratic  form  if  it  contains 
but  two  powers  of  the  unknown  number,  and  the  exponent 
of  one  power  is  exactly  twice  that  of  the  other  power. 

302.  Equations  in  the  quadratic  form  may  be  solved  by 
the  methods  for  solving  quadratics. 

1.  Solve  8  a6  +  63  x*  =  8. 

Multiply  by  32  and  complete  the  square, 

256  a;6 +  (    )  +  63*  =  4225. 
Extract  the  square  root,  16  x3  +  63  =  ±  65. 

Hence,  x8  =  -»  or  —  8. 

o 

Extracting  the  cube  root,  two  values  of  x  are  i  and  —  2.  There 
are  four  other  values  of  x  which  may  be  found  by  §  297. 

2.  Solve  Vx1  -  3^/x8  =  40. 
Using  fractional  exponents  we  have, 

x*-3x*  =  40. 
Complete  the  square,       4  x*  —  12  x*  +  9  =  169. 


Extract  the  root, 

2x^-3  =  ±13. 

Transpose  —  3, 

2xJ  =  16,  or —  10. 

Divide  by  2, 

x*  =  8,  or  —  5. 

Extract  the  cube  root, 

x*  =  2,  or  —  5*. 

Raise  to  the  fourth  power, 

x  —  16,  or  5VH. 

3.   Solve  (2x-Sy-  (2x 

-  3)  =  6. 

Put  y  for  2  x  —  3,  and  therefore  y2  for  (2  x  —  3)2. 

We  have 

y2-y  =  G. 

Solving, 

y  =  3,  or  —  2. 

Put2x  — 3fory, 

2x-3  =  3, 

2x-3  =  ~ 

,  *  =  3. 

x  =  i 

282  QUADRATIC  EQUATIONS. 

4.  Solve  7a2-5a  +  8V7a2-5a  +  l  =  -8. 

By  adding  1  to  both  sides,  we  have 

7x2-5x  +  l  +  8V7x2-5x  +  l  =  -7. 
Put  y  for  V7x2  —  5x  +  1,  and  hence  y2  for  7 x2  —  5*  +  1. 
Then,  y2  +  8y  =  -7. 

Solving,  y  =  -  l,  or  —  7. 

Therefore,  y2  =  1,  or  49. 

We  now  have  7x2  —  5x  +  1  =  1,  or  7x2  —  5x  +  1  =  49. 

Solving  these  equations,  we  find  for  the  values  of  x, 

0,§;or3,-f 

These  values  all  satisfy  the  given  equation  when  we  take  the  nega- 
tive value  of  the  square  root  of  the  expression  7  x2  —  5  x  +  1 ;  they 
are  in  fact  the  four  roots  of  the  biquadratic  obtained  by  clearing  the 
given  equation  of  radicals. 

5.  Solve  x*  -  10a8  +  35 a2-  50a:  +  24  =  0. 
Take  the  square  root  of  the  left  side. 

a4  -  10a8  +  35  a2  -  50  a  +  24|a2_-5a_+5 


2a2-5a 


-  10a3  +  35  a2 

-  10a8  +  25  a2 


2a2 -10a +  5 


10a2 -50a +  24 
10a2 -50a +  25 


It  is  now  seen  that  if  1  is  added,  the  square  will  be  complete  and 
the  equation  will  be 

a4  -  10a8  +  35a2  -  50a  +  25  =  1. 

Extract  the  square  root,  and  the  result  is 

x2-5x  +  6  =  ±l. 
Solving,  x  =  4,  1,  3,  or  2. 


QUADRATIC  EQUATIONS.  283 

Exercise  117.    . 
Solve: 

1.  a;4-5a;2  +  4  =  0.  6.  10  x*  -  21  =  x\ 

2.  x4-  13a;2  +  36  =  0.  7.  -</tf2  +  3Vz  =  lf. 

3.  a;4 -21  a;2  =  100.  8.  3Va-2Va^  =  -20. 

4.  4a6 -3a-8  =  27.  9.  5 x*»  +  3 xn  =  6£ . 

5.  2a;4  +  5a;2  =  21£.  10.  (8a;  +  3)2+(8a;  +  3)=30. 


11.     2(«2-£C  +  l)-ViC2-iC  +  l  =  l. 


12.   a;6-9a:8  +  8  =  0.  14.    (x  +  1)  +  Va;  +  1  =  6. 

13 
12.    a*4-ic-^  =  ~-  15.   a;4 -13 a;2  =-36. 

O 


16.    2a;2  +  4a;  +  9  +  3V2a:2  +  4a;  +  9  =  40. 


17.    2a;2  +  3a;-5V2a;2  +  3a;  +  9  =  -3. 


18.    3a;2  +  15a;-2Va;2  +  5a;  +  l  =  2. 


19.   x2-§x  +  3y/2x2-3x  +  2  =  7. 


20.   2  a;2 -Va;2  -  2a;  -  3  =  4a;  +  9. 


21.    3a;2  -  4a;  +  V3a;2  -  4a;  -  6  =  18. 


22.  3a;2-7  +  3V3a;2-16a;  +  21  =  16x. 

23.  a;4  -  2a;8  -  13a;2  +  14a;  +  24  =  0. 

24.  a;4  -  4a;8  -  10a;2  +  28a; -15  =  0. 

25.  4a;4-20a;8  +  23a;2  +  5a;-6  =  0. 

26.  4a;4  -  12a;8  +  5a;2  +  6x  -15  =  0. 

27.  4a;4  -  12a;8  +  17a;2 -12a; -12  =  0. 

28.  6sc2  +  6a;  +  Va;(x  +  1)  =  7, 


284  QUADRATIC  EQUATIONS. 

Character  of  the  Roots. 

The  two  roots  of  ax2  +  bx  +  c  =  0  are 


b     ,    Wb^-Tac       ,         b        ^/b2-±ac 

^-  +  ^~5 and-- - (§299) 

2a  2a  2a  2a  v  ' 


303.  The  character  of  the  two  roots  depends  wholly  upon 
b2  —  4  ac,  the  expression  under  the  radical  sign. 

1.  If  b2  —  4  ac  is  positive  and  not  zero,  the  roots  are  real 
and  unequal. 

The  roots  are  real,  since  the  square  root  of  the  positive 
number,  62  —  4  ac,  can  be  found  exactly  or  approximately. 

The  roots  are  unequal,  since  V#2  —  4  ac  is  not  zero. 

If  b2  —  4  ac  is  a  perfect  square,  the  roots  are  rational;  if 
b2  —  4  ac  is  not  a  perfect  square,  the  roots  are  surds. 

2.  If  b2  —  4  ac  is  zero,  the  two  roots  are  real  and  equal, 

since  they  both  become  —  —  •     Hence, 

2  a 

The  roots  of  ax2  -f  foe  +  c  =  0  are  re<z£,  if  #2  =  or  >  4  ac, 


3.  If  b2  —  4  ac  is  negative,  the  roots  are  imaginary,  since 
they  involve  the  square  root  of  a  negative  number.     Hence, 

The  roots  of  ax2  -f  bx  -f  c  =  0  are  imaginary,  if  £2  <  4  ac. 

The  two  imaginary  roots  of  a  quadratic  cannot  be  equal, 
since  b2  —  4  ac  is  not  zero.  They  have,  however,  the  same 
real  parts,  and  the  same  imaginary  parts  with  opposite  signs, 
and  are,  therefore,  conjugate  imaginaries,  §  284. 

The  expression  b2  —  4  ac  is  called  the  discriminant  of  the 
expression  ax2  +  bx  -f-  c. 


QUADRATIC  EQUATIONS.  285 

304.  The  above  cases  may  be  summarized  as  follows : 

Case  1.    If  b2  —  4  ac  >  0,  the  roots  are  real  and  unequal. 
Case  2.    If  b2  —  4  ac  =  0,  the  roots  are  real  and  equal. 
Case  3.    If  b2  —  4  ac  <  0,  the  roots  are  imaginary. 

305.  By  finding  the  value  of  b2  —  4  ac  we  can  determine 
at  once  the  character  of  the  roots  of  a  given  equation. 

1.  x2-  5a  +  6  =  0. 

Here  a  =  1,  b  —  —  5,  c  =  6. 

&2  -  4  ac  =  25  -  24  =  1. 
The  roots  are  real  and  unequal,  and  rational. 

2.  3a2  +  7a:-l  =  0. 

Here  a  =  3,  b  =  7,  c  =  —  1. 

62  _  4  ac  =  49  +  12  =  61. 

The  roots  are  real  and  unequal,  and  are  both  surds. 

3.  4a2-12z  +  9  =  0. 

Here  a  =  4,  b  =  —  12,  c  =  9. 

62  -  4  ac  =  144  -  144  =  0. 
The  roots  are  real  and  equal. 

4.  2x2-  Sx  +  4  =  0. 

Here  a  =  2,  b  =  -  3,  c  =  4. 

62  —  4  ac  =  9  —  32  =  -  23. 
The  roots  are  both  imaginary. 

5.  Find  the  values  of  m  for  which 

2mx2+  (5m4-2)«  +  (4m  +  l)  =0 
has  its  two  roots  equal . 

Here      •  a  =  2m,6  =  5m  +  2,c  =  4m  +  l. 

If  the  roots  are  to  be  equal,  we  must  have 

62  —  4ac  =  0,  or  (5m +  2)2  —  8m(4m  +  l)=0. 

2 
Solving,  m  =  2,  or  —  -  • 

s 


286  QUADRATIC  EQUATIONS. 

For  these  values  of  m  the  equation  becomes 

4x*  +  12x  +  9  =  0,  and  4x2  —  4x  +  1  =0, 
each  of  which  has  its  roots  equal. 

Exercise  118. 

Determine  witliout  solving  the  character  of  the  roots  of 
each  of  the  following  equations : 

1.  jc2  +  5a;  +  6  =  0.  6.  6x2-7x-3  =  0. 

2.  a2  +  2a; -15  =  0.  7.  5x2  -  5x  -  3  =  0. 

3.  x2  +  2x  +  3  =  0.  8.  2x2-x  +  5  =  0. 

4.  3  a;2 +  7  a; +  2  =  0.  9.  6a2  +  a  -  77  =  0. 

5.  9a;2  +  6a; +  1  =  0.  10.    5a;2  +  8a;  +  ^  =  0. 

5 

Determine  the  values  of  m  for  which  the  two  roots  of 
each  of  the  following  equations  are  equal : 

11.  (m  +  l)a;2  +  (m-l)a;  +  ra  +  l  =  0. 

12.  (2  m  -  3)  x2  +  mx  +  m  -  1  =  0. 

13.  2mx2  +  a:2  +  4a;  +  2ma;-|-2m-4  =  0. 

14.  2 mx2  +  3 ma;  —  6  =  Sx  —  2  m  —  x2. 

15.  mx2 +  9x  -  10  =  3mx-2x2  + 2m. 

Relations  of  Roots  and  Coefficients. 

306.    If  we  divide  the  general  equation  ax2  +  bx  +  c  —  0 

b         c 
by  a,  we  have  the  equation  a;2  +  -a;  +  -  =  0;   this  may  be 

be 
written  x2  +  jpx  -\-  q  =  0,  where  p  =  ->  ^  =  -  • 

ot  At 


QUADRATIC  EQUATIONS.  287 

307.    By  solving  x2  -f-  px  -f-  q  =  0,"and  denoting  the  first 
value  of  x  by  rlf  and  the  second  value  by  r2,  we  have 

1  2^         2 


r*~      2  2 

Add,  rj  +  r2  =  —  ^?. 

Multiply,  r\Tt  =  y. 

It  appears,  then,  that  if  any  quadratic  equation  is  made 
to  assume  the  form  x2  -{-px  -f-  q  =  0,  the  following  relations 
hold  between  the  coefficients  and  roots  of  the  equation : 

1.  The  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  x  with  its  sign  changed. 

2.  The  product  of  the  two  roots  is  equal  to  the  constant 
term. 


308.  If  rx  and  r2  are  the  roots  of  the  equation  x2  -{-px 
+  q  =  0,  the  equation  may  be  written 

(x  -  rj)  (x  -  r2)  =  0. 

309.  Form  the  equation  of  which  the  roots  are  3  and  —  2. 

The  equation  is  (x  —  3)  (x  +  2)  =  0, 

or  x2  —  x  —  6  =  0. 


Exercise  119. 
Form  the  equation  of  which  the  roots  are : 
1.    7,  6.  5.    ljfc  -  1J.  9.    3  +  V2,  3  -  V2. 


2.    5,-3. 

6.    -li,-lf. 

10. 

l  +  V^i ,  l - 

-V=I 

3.    l*,-2. 

7.    13,-4*. 

n. 

a,  a  —  b. 

4.    4,  2J. 

ft      3      2 

8*  IT  II 

12. 

a  -{-  b,  a  —  b. 

288  QUADRATIC  EQUATIONS. 

Problems  Involving  Quadratics. 

310.  Problems  that  involve  quadratic  equations  appar- 
ently have  two  solutions,  since  a  quadratic  equation  has 
two  roots.  If  both  roots  of  the  quadratic  equation  are 
positive  integers,  they  will,  generally,  both  be  admissible 
solutions. 

Fractional  and  negative  roots  will  in  some  problems  give 
admissible  solutions ;  in  other  problems  they  will  not. 

No  difficulty  will  be  found  in  selecting  the  result  which 
belongs  to  the  particular  problem  we  are  solving.  Some- 
times, by  a  change  in  the  statement  of  the  problem,  we  may 
form  a  new  problem  which  corresponds  to  the  result  that 
was  inapplicable  to  the  original  problem. 

Imaginary  roots  indicate  that  the  problem  is  impossible. 

Here,  as  in  simple  equations,  x  stands  for  an  unknown 
number. 

1.  The  sum  of  the  squares  of  two  consecutive  numbers 
is  481.     Find  the  numbers. 

Let  x  —  one  number, 

and  x  +  1  =  the  other. 

Then,  x2  +  (x  +  l)2  =  481, 

or  2x2  +  2x  +  l  =481. 

The  solution  of  which  gives  x  =  15,  or  —  16. 

The  positive  root  15  gives  for  the  numbers,  15  and  16. 

The  negative  root  —  16  is  inapplicable  to  the  problem,  as  consecu- 
tive numbers  are  understood  to  be  integers  which  follow  one  another 
in  the  common  scale,  1,  2,  3,  4  

2.  A  pedler  bought  a  number  of  knives  for  $2.40.  Had 
he 'bought  4  more  for  the  same  money,  he  would  have  paid 
3  cents  less  for  each.  How  many  knives  did  he  buy,  and 
what  did  he  pay  for  each  ?  * 

Let  x  =  number  of  knives  he  bought. 

240 
Then,  =  number  of  cents  he  paid  for  each 


QUADRATIC  EQUATIONS.  289 


But  if  x  +  4  =  number  of  knives  he  bought, 

— —  =5  number  of  cents  he  paid  for  each, 
x  +  4 


—7  =  the  difference  in  price. 

x       x  +  4 

But 

3  =  the  difference  in  price. 

240        240 
x       x  +  4 

Solving, 

x  =  16,  or  —  20. 

He  bought  16  knives,  therefore,  and  paid  *$*,  or  15 
cents  for  each. 

If  the  problem  is  changed  so  as  to  read :  A  pedler  bought 
a  number  of  knives  for  $ 2.40 ;  if  he  had  bought  4  less 
for  the  same  money,  he  would  have  paid  3  cents  more  for 
each,  the  equation  will  be 

240        240  _  _ 

- —  —  o. 


x  —  4        x 
Solving,  x  =  20,  or  —  16. 

This  second  problem  is  therefore  the  one  which  the  nega- 
tive answer  of  the  first  problem  suggests. 

3.   What  is  the  price  of  eggs  per  dozen  when  2  more  in 
a  shilling's  worth  lowers  the  price  1  penny  per  dozen  ? 

Let  x  =  number  of  eggs  for  a  shilling. 

Then,  -  =  cost  of  1  egg  in  shillings, 

x 

12 
and  —  =  cost  of  1  dozen  in  shillings. 

But  if  x  +  2  =  number  of  eggs  for  a  shilling, 

12 

=  cost  of  1  dozen  in  shillings. 

X  T  £i 

12         12  1 

•*•  —  -  J-^2  =  12  (1  penny  being  *  °f  a  shillinS)' 

The  solution  of  which  gives  x  =  16,  or  —  18. 

And,  if  16  eggs  cost  a  shilling,  1  dozen  will  cost  9  pence. 

Therefore,  the  price  of  the  eggs  is  9  pence  per  dozen. 


290  QUADRATIC  EQUATIONS. 

If  the  problem  is  changed  so  as  to  read :  What  is  the 
price  of  eggs  per  dozen  when  two  less  in  a  shilling's  worth 
raises  the  price  1  penny  per  dozen  ?  the  equation  will  be 

12        12  _  1 
x  —  2       x       12 ' 
The  solution  of  which  gives  x  =  18,  or  —  16. 

Hence,  the  number  18,  which  had  a  negative  sign  and  was  inappli- 
cable in  the  original  problem,  is  here  the  true  result. 

Exercise  120. 

1.  The  sum  of  two  squares  of  two  consecutive  integers  is 
761.     Find  the  numbers. 

2.  The  sum  of  the  squares  of  two  consecutive  numbers  ex- 
ceeds the  product  of  the  numbers  by  13.    Find  the  numbers. 

3.  The  square  of  the  sum  of  two  consecutive  even  num- 
bers exceeds  the  sum  of  their  squares  by  336.  Find  the 
numbers. 

4.  Twice  the  product  of  two  consecutive  numbers  ex^ 
ceeds  the  sum  of  the  numbers  by  49.     Find  the  numbers. 

5.  The  sum  of  the  squares  of  three  consecutive  numbers 
is  110.     Find  the  numbers. 

6.  The  difference  of  the  cubes  of  two  consecutive  odd 
numbers  is  602.     Find  the  numbers. 

7.  The  length  of  a  rectangular  field  exceeds  its  breadth 
by  2  rods.  If  the  length  and  breadth  of  the  field  were 
each  increased  by  4  rods,  the  area  would  be  80  square  rods. 
Find  the  dimensions  of  the  field. 

8.  The  area  of  a  square  may  be  doubled  by  increasing 
its  length  by  10  feet  and  its  breadth  by  3  feet.  Find  the 
length  of  its  side. 


QUADRATIC  EQUATIONS.  291 

9.  A  rectangular  grass  plot  12  yards  long  and  9  yards 
wide  has  a  path  around  it.  The  area  of  the  path  is  §  of 
the  area  of  the  plot.     Find  the  width  of  the  path. 

10.  The  perimeter  of  a  rectangular  field  is  60  rods.  Its 
area  is  200  square  rods.     Find  its  dimensions. 

11.  The  length  of  a  rectangular  plot  is  10  rods  more 
than  twice  its  width,  and  the  length  of  a  diagonal  of  the 
plot  is  25  rods.     What  are  the  dimensions  of  the  plot  ? 

12.  The  denominator  of  a  certain  fraction  exceeds  the 
numerator  by  3.  If  both  numerator  and  denominator  are 
increased  by  4,  the  fraction  will  be  increased  by  J.  Find 
the  fraction. 

13.  The  numerator  of  a  fraction  exceeds  twice  the  de- 
nominator by  1.  If  the  numerator  is  decreased  by  3,  and 
the  denominator  increased  by  3,  the  resulting  fraction  will 
be  the  reciprocal  of  the  given  fraction.     Find  the  fraction. 

14.  A  farmer  sold  a  number  of  sheep  for  $120.  If  he 
had  sold  5  less  for  the  same  money,  he  would  have  received 
$2  more  a  sheep.     How  much  did  he  receive  a  sheep  ? 

State  the  problem  to  which  the  negative  solution  applies. 

15.  A  merchant  sold  a  certain  number  of  yards  of  silk 
for  $40.50.  If  he  had  sold  9  yards  more  for  the  same 
money,  he  would  have  received  75  cents  less  per  yard. 
How  many  yards  did  he  sell  ? 

16.  A  man  bought  a  number  of  geese  for  $27.  He  sold 
all  but  two  for  $25,  thus  gaining  25  cents  on  each  goose 
sold.     How  many  geese  did  he  buy  ? 

17.  A  man  agrees  to  do  a  piece  of  work  for  $48.  It 
takes  him  4  days  longer  than  he  expected,  and  he  finds 
that  he  has  earned  $1  less  per  day  than  he  expected.  In 
how  many  days  did  he  expect  to  do  the  work  ? 


292  QUADRATIC  EQUATIONS. 

18.  Find  the  price  of  eggs  per  dozen  when  10  more  in 
one  dollar's  worth  lowers  the  price  4  cents  a  dozen. 

19.  A  man  sold  a  horse  for  $171,  and  gained  as  many 
per  cent  on  the  sale  as  the  horse  cost  dollars.  How  much 
did  the  horse  cost  ? 

20.  A  drover  bought  a  certain  number  of  sheep  for  $160. 
He  kept  four,  and  sold  the  remainder  for  $10.60  per  head, 
and  made  on  his  investment  f  as  many  per  cent  as  he  paid 
dollars  for  each  sheep  bought.    How  many  sheep  did  he  buy  ? 

21.  Two  pipes  running  together  can  fill  a  cistern  in  5| 
hours.  The  larger  pipe  will  fill  the  cistern  in  4  hours  less 
time  than  the  smaller.  How  long  will  it  take  each  pipe 
running  alone  to  fill  the  cistern  ? 

22.  A  and  B  can  do  a  piece  Of  work  together  in  18  days, 
and  it  takes  B  15  days  longer  to  do  it  alone  than  it  does  A. 
In  how  many  days  can  each  do  it  alone  ? 

23.  A  boat's  crew  row  4  miles  down  a  river  and  back 
again  in  1  hour  and  30  minutes.  Their  rate  in  still  water 
is  2  miles  an  hour  faster  than  twice  the  rate  of  the  current. 
Find  the  rate  of  the  crew  and  the  rate  of  the  current. 

24.  A  number  is  formed  by  two  digits.  The  units'  digit 
is  2  more  than  the  square  of  half  the  tens'  digit,  and  if  18 
is  added  to  the  number,  the  order  of  the  digits  will  be- 
reversed.     Find  the  number. 

25.  A  circular  grass  plot  is  surrounded  by  a  path  of  a 
uniform  width  of  3  feet.  The  area  of  the  path  is  $  the  area 
of  the  plot.     Find  the  radius  of  the  plot. 

26.  If  a  carriage  wheel  11  feet  round  took  \  of  a  second 
less  to  revolve,  the  rate  of  the  carriage  would  be  five  miles 
more  per  hour.     At  what  rate  is  the  carriage  traveling  ? 


CHAPTER  XX. 
SIMULTANEOUS  QUADRATICS. 

311.  Quadratic  equations  involving  two  unknown  num- 
bers require  different  methods  for  their  solution,  according 
to  the  form  of  the  equations. 

Case  1. 

312.  When  one  of  the  equations  is  a  simple  equation. 

Solve  Sx2-  2x7/ =  5^  (1) 

x-y  =  2  J  (2) 

Transpose  x  in  (2),  y  =  x  —  2. 

In  (1)  put  x  —  2  for  y, 

3x2  —  2x(x  —  2)  =  5. 
The  solution  of  which  gives        x  =  1,  or  x  =  —  5. 
If  x  =  1, 

y  =  l-2  =  -l;    , 
and  if  a  =  —  5, 

y  =  -5-2  =  -7. 
We  have,  therefore,  the  pairs  of  values, 

x  =  l      I  x  =  —  5 

1  (  J  or  » 

y  —  — 1  )  '       y  —  —  l 

The  original  equations  are  both  satisfied  by  either  pair  of  values, 
But  the  values  x  =  l,y  =  —  7,  will  not  satisfy  the  equations ;  nor  will 
the  values  x  =  —  5,  y  =  —  1. 

The  student  must  be  careful  to  join  to  each  value  of  x 
the  corresponding  value  of  y. 


294 


SIM UL TANEOUS   Q UADRA TICS. 


Case  2. 

313.  When  the  left  side  of  each  of  the  two  equations  is  homo- 
geneous and  of  the  second  degree. 


Solve 


2y2  -  Axy  +  Sx2  =  17 1 


y2-x2  =  16 


Let  y  =  vx  and  substitute  vx  for  y  in  both  equations. 
From  (1),  2  v*x2  —  4  ux2  +  3  x2  =  17. 

r.2  — 


17 


.-.  x* 


From  (2), 


2t>2  — 4t>  +  3 
tj2x2  —  x2  =  16. 
16 


.-.  x2  = 


v2—  1 


17  1ft 

Equate  the  values  of  x2,  — = — r-s  as  — — -» 

2  v2  —  4  v  +  3      u2  —  1 

32«2-64u  +  48  =  17t>2-17, 

15  u2  —  64  u  =  —  65, 

225u2-960u  =  -976, 

225u2-()  +  322  =  49, 

15v-32  =  ±7. 


If 


y  =VX  = 


5x 


Substitute 

in  (2), 

25  x2 

9 

-x2  = 

16, 

x2  = 

9, 

x  = 

±3, 

V~ 

5x 
'3 

±5. 

5 

>  =  -,  or 

13 . 
5  ' 

If 

V 

13 

y 

=  VX 

_13x% 
6 

Substitute  in 

(2), 

169x2 
25 

—  X2 

=  16, 

X2 

25 

X 

=4 

y  = 

13  x 
5 

-*¥ 

(1) 

(2) 


SIMULTANEOUS   QUADRATICS. 


295 


Case  3. 

314.   When  the  two  equations  are  symmetrical  with  respect 
to  x  and  y ;  that  is,  when  x  and  y  are  similarly  involved. 

Thus,  the  expressions 

2x»  +  3«V  +  2ys,  2xy-3x-3y  +  l,x*-  3x2y  -3xy2  +  y4 

are  symmetrical  expressions.  In  this  case  the  general  rule 
is  to  combine  the  equations  in  such  a  manner  as  to  remove 
the  highest  powers  of  x  and  y. 


Solve 


x*  +  ^  =  3371 
«.+Jf  =      7J 


To  remove  x4  and  y4,  raise  (2)  to  the  fourth  power, 
Add  (1), 


x4  +  4x32/  +  6x22/2  +  4x2/8  +    2/4  =  2401 
x*  +    2/4=    337 


2x4  +  4x3j/  +  6xV  +  4xy3  +  22/4  =  2738 
Divide  hy  2,  x4  +  2  x32/  +  3  xfy2  +  2  x?/8  +  2/4  =  1369. 
Extract  the  square  root,  x2  +  xy  +  y2  =  ±  37. 

Subtract  (3)  from  (2)2,      xy  =  12  or  86. 
We  now  have  to  solve  the  two  pairs  of  equations, 


From  the  first, 


From  the  second, 


a  +  y  =    7)     x  +  y  =   7) 
xy  =  12  J  '        xy  =  86  ) ' 


2/  =  3f;°r2/  =  4} 


7  ±  V-  295 


v  = 


7qzV-295 


(1) 

(2) 


(3) 


296  SIMULTANEOUS   QUADRATICS. 

315.  The  preceding  cases  are  general  methods  for  the 
solution  of  equations  that  belong  to  the  kinds  referred  to ; 
often,  however,  in  the  solution  of  these  and  other  kinds  of 
simultaneous  equations  involving  quadratics,  a  little  inge- 
nuity will  suggest  some  step  by  which  the  roots  may  be 
found  more  easily  than  by  the  general  method. 

1.    Solve  x  +  y  =  40  T  (1) 

xy  =  300     J  (2) 

Square  (1),  x2  +  2  xy  +  y*  =  1600.  (3) 

Multiply  (2)  by  4,  4xy  =  1200.  (4) 

Subtract  (4)  from  (3), 

x2-2xy  +  ?/2  =  400.  (5) 

Extract  the  square  root,  x  —  y  =  ±  20.  (6) 


From  (1)  and  (6),                              x  =  30  )  . 

y  =  io  r 

x  =  10 
%  =  30 

I; 

I  1        9 

2.   Solve                          £  +  £  =  ^ 
jc      ?/      20 

II  41 

arJ~*V~400J 

' 

(i) 

(2) 

c            ™                       *  j_  2  a.  *  -  81 
Square(l),                   _+-  +  ___. 

(3) 

2        40 

Subtract  (2)  from  (3),                   —  =  ^  ■ 

W 

Subtract  (4)  from  (2), 

J___2      i_j_. 
x2      xy      y2      400 

Extract  the  square  root,         -  — =  ±  —  • 
x      y          £\j 

(6) 

From  (1)  and  (6),                           x  =  4  i  ;  or 

y  =  5  >  ' 

x  =  5)  t 

y  =  4i  ' 

SIMULTANEOUS   QUADRATICS.  297 


x   —  y  —    41 
x2  +  y2  =  40  J 


3.    Solve  x  -y  =    4\  (1) 

(2) 


Square  (1),  x2  -  2  xy  +  y2  =  16.  (3) 

Subtract  (2)  from  (3),     —  2xy  =  -  24.  (4) 

Subtract  (4)  from  (2), 

x2  +  2  xy  +  y2  =  64. 
Extract  the  root,  x  +  y  =  ±  8.  (5) 

From  (1)  and  (5),  x  =  6)  x=  ~  2  ) 

y  =  2)  ;  °r  y=~6J 

4.    Solve  a?  +  y*  =  91\  (1) 

«:  +  Jf  =    7  J  (2) 

Divide  (1)  by  (2),  x2  -  xy  +  y2  =  13.  (3) 

Square  (2),  x2  +  2xy  +  y2  =  49.  (4) 

Subtract  (3)  from  (4),         3xy  =  36. 

Divide  by  —  3,  —  xy  —  — 12.  (6) 

Add  (5)  and  (3),  x2  —  2  xy  +  y2  =  1. 

Extract  the  root,  x  —  y  =  db  1.  (6) 

From  (2)  and  (6),  x  =  4  )  a;  =  3  ) 

„=3J;  °r  y=4> 


5.    Solve                  «,  +  y,  =  18ajy1  (1) 

a;  +y  =12       J  (2) 

Divide  (1)  by  (2),  a2  -  xy  +  y2  =  $&•  •  (3) 

Square  (2),         x2  +  2xy  +  y2  =  144.  (4) 
Subtract  (4)  from  (3),     —  Sxy  =  -|*  -  144, 


which  gives  xy  =  32 

ry=  12 
xy  =  32 
x  =  8| 
y  =  4  J  '        y 


We  now  have,  x  +  y  =  12  ) 

Solving,  we  find,  x  =  8^  x  =  4 ) 


298 


SIM UL TANE OUS    Q UADEA TICS. 


Exercise  121. 

Solve : 

1.    aj  +  y  =  7l          3. 

x  - 

-y  =  6) 

5.    x  +  y  =  12 

xy  =  10     J 

xy 

=  -*} 

X2  +  y2  =  80 

2.    a  +  y  =  12|        4. 

x  - 

-y  =  10 

1         6.    a;  4  2/ =  3 

a;y  =  27       J 

xy 

=  11 

J               a;2  +  ?/2  =  29 

7.    x   -y  =    9l 

17. 

a;2-^2  =  9l 

a;2  +  y2  =  45j 

a;  -y  =lj 

8.   a  +  2y  =    71 

18. 

a;2  +  3?/  +  17  =  0J 

a2  +     2/2  =  10  J 

3a;  —  y  =  3 

9.    3a;  -y  =12l 

Z2   _^2=16J 

10.  y  =  3x  +  l    1 
a;2  +  a;y  =  33J 

11.  5a;  —  4>  =  10*1 

19. 
20. 

a;      y 

1        1       / 

a;2      ?/2 

a;      y      6 

3«a-42/2=    8J 
12.   x  +  7y  =  2S'\ 

11       13 

a;2  +  y2      36 

xy  =  6           J 

21. 

3a;  +  2y  =  2a-?/j 

13.  2«  -3y    =2 

a;2-2a;2/=-7 

14.  2a;   -Sy    =    l] 
3a;2 -4a;*/ =  32  J 

} 

22. 

xy  =  6 

l+l=ii 

a;2      y* 

J 

15.    x2-xy  +  y2  =  21 

•1 

23. 

8a;  +  6y  =  4a;yj 

x  +  y  =  9 

J 

a;y  =  16 

16.    a;2-3a;y  +  22/2  = 

.01 

24. 

a;»  +  2/8  =  35\ 

2a;  +  3y  =  7 

J 

a;  +y  =    5 

J 

SIMULTANEOUS   QUADRATICS. 


299 


Z'l} 


25.  x8  —  y8 

x    -y  = 

26.  x*  +  y8  =  65 
#  4-  y  =    5 

27.  a2?/  +  ay2  =  120 1 
x  +  y  =  8 

28.  aj«~y»  =  ^' 
a  -y  =  i 

29.  a8-fy8  =  126 

x2  —  xy  +  y2  =  21 


28 


30. 

a?3  —  t/3  =  56 

a;2  +  xy  -f-  y2  = 

31. 

^      ?^_35" 
y       x       3 

«     y.    12 

- 

32. 

.a2      y2  _  19  ' 
y       x        2 
1       11 

y     #"18^ 

- 

33. 

z2  +a;y  =  24l 

#y  +  y2  =  40  J 

3*4. 

x2  —  xy  =  8  ^ 

ay  —  y2  =  7 

35.    z2  +  2zy  =  24      1 
2ay  +  4y2  =  120J 


36.    ±x2  +5 
7*y4-9 


xy  =  14  1 
y2=50j 


37.    a2 +  a;y  +  y2  =  39 
2x2  +  3xy  +  y2  =  63 


38. 

x*  -f  3y2  =  52 

ay  +  2  y2  =  40 

39. 

2  a2  -  y2  =  46 ' 

xy    +  y2  =  14 

40.    cc2  +  scy  +  2y2  = 
2cc2-3zy  +  2 


=  44       1 

y2  =  16j 


41. 

x*  +  3y2  =  31 

±xy  +  y2  =  33 

42. 

3x*+7xy  =  8 

x2  4-  5  xy  4-  9  y 

43. 

a44-y4  =  97J 

a;  4-y  =    5J 

44. 


*44-y4  =  17J 
^  4-y  =    3  J 


45.   z4  +  y4  =  881*1 
x  -y  =      1J 


;} 


46.  a6  +  y6  =  211 

x   4-y 

47.  a5-y5  =  242*1 
x   -y  =      2  J 


48.  a;2  4-  y2  =  #y 
a;  4-y  =  ay 


.    a;8  —  y3  =  7  ay  1 
x  -y  =2      J 


300 


SIMUL TANEOUS    Q UADRA TICS. 
61 


60.    x*  +  y*  =  36xy]  61.    x*-y8  =  a8\ 

x  +y  =  24       J  x   -  y  =  a  ] 


61.    x8  +  3xy2  =  62 
3x2y  +  y8  =  63 


52.    x2+  xy  +y2  =      61 1 
(c4  +  £cV+2/4  =  1281J 


53.    a;2-  xy  +  y2  =    3 
a;4  +  x2y2  +  y4  =  21 


54.  x+v  |  ^  -  y  __  10 


#•*  i-  y2 


2/     «  +y 

20 


55, 


x  —  y  x  -f-  y  _  24 
x  -\-  y  x  —  y  5 
3a  +  4y  =  36 


56.  x2  +  y2  +  x  +  y  = 
xy  +  l(S  =  0 

57.  a?  -  y  -  3  =  0 
2(x2-y2)  =  3xy 


32 


58.    i  +  i  =  7 

x       y 


x+1       y  +  1 


31 

20 


59.    a4-f  y4  =  272 

#2  +  y2  —  3  xy  —  4 


62.  *  +  f  =  l 

a       b 

as      y 

63.  a?2  =  ax  4"  6y 

y2  =  bx  +  ay 

64.  a2 -f  y2  =  2  (a2 + 
xy  =  a2  —  b2 

a4  +  b* 


65.    x2  +  2/2  = 


aw 


66.    x2-y2  = 


a2b2 


a  —  b 


a  +  b 


xy  = 


ab 


(a  +  by 


} 


67.    x2  —  xy  = 

ay  -  y2 


=  2a&  +  2&2] 
=  2a&-2&2J 


68.    x2  — 
xy  =  b 


tf  =  a* 


69.    x2-y2  =  8ab^ 
xy  =  a2—  4:b2  J 


60.    x2  +  2/2  =  £cy +  ll  70.    ic8  +  y8  =  a8  +  58l 

«+y=2icy  —  lj  x   +  y  =  a  +b  \ 


SIMULTANEOUS  QUADRATICS.  301 

Exercise  122. 

1.  The  area  of  a  rectangle  is  60  square  feet,  and  its 
perimeter  is  34  feet.  Find  the  length  and  breadth  of  the 
rectangle. 

2.  The  area  of  a  rectangle  is  108  square  feet.  If  the 
length  and  breadth  of  a  rectangle  are  each  increased  by 
3  feet,  the  area  will  be  180  square  feet.  Find  the  length 
and  breadth  of  the  rectangle. 

3.  If  the  length  and  breadth  of  a  rectangular  plot  are 
each  increased  by  10  feet,  the  area  will  be  increased  by  400 
square  feet.  But  "if  the  length  and  breadth  are  each  dimin- 
ished by  5  feet,  the  area  will  be  75  square  feet.  Find  the 
length  and  breadth  of  the  plot. 

4.  The  area  of  a  rectangle  is  168  square  feet,  and  the 
length  of  its  diagonal  is  25  feet.  Find  the  length  and 
breadth  of  the  rectangle. 

5.  The  diagonal  of  a  rectangle  is  25  inches.  If  the 
rectangle  were  4  inches  shorter  and  8  inches  wider,  the 
diagonal  would  still  be  25  inches.  Find  the  area  of 
the  rectangle. 

6.  A  rectangular  field,  containing  180  square  rods,  is 
surrounded  by  a  road  1  rod  wide.  The  area  of  the  road  is 
58  square  rods.     Find  the  dimensions  of  the  field. 

7.  Two  square  gardens  have  a  total  surface  of  2137 
square  yards.  A  rectangular  piece  of  land  whose  dimen- 
sions are  respectively  equal  to  the  sides  of  the  two  squares 
will  have  1093  square  yards  less  than  the  two  gardens 
united.     What  are  the  sides  of  the  two  squares  ? 

8.  The  sum  of  two  numbers  is  22,  and  the  difference  of 
their  squares  is  44.     Find  the  numbers. 


302  SIMULTANEOUS   QUADRATICS. 

9.    The  difference  of  two  numbers  is  6,  and  their  product 
exceeds  their  sum  by  39.     Find  the  numbers. 

10.  The  sum  of  two  numbers  is  equal  to  the  difference 
of  their  squares,  and  the  product  of  the  numbers  exceeds 
twice  their  sum  by  2.     Find  the  numbers. 

11.  The  sum  of  two  numbers  is  20,  and  the  sum  of  their 
cubes  is  2060.     Find  the  numbers. 

12.  The  difference  of  two  numbers  is  5,  and  the  differ- 
ence of  their  cubes  exceeds  the  difference  of  their  squares 
by  1290.     Find  the  numbers. 

13.  A  number  is  formed  of  two  digits..  The  sum  of  the 
squares  of  the  digits  is  58.  If  twelve  times  the  units' 
digit  is  subtracted  from  the  number,  the  order  of  the  digits 
will  be  reversed.     Find  the  number. 

14.  A  number  is  formed  of  three  digits,  the  third  digit 
being  twice  the  sum  of  the  other  two.  The  first  digit  plus 
the  product  of  the  other  two  digits  is  25.  If  180  is  added 
to  the  number,  the  order  of  the  first  and  second  digits  will 
be  reversed.     Find  the  number. 

15.  There  are  two  numbers  formed  of  the  same  two 
digits  in  reverse  order.  The  sum  of  the  numbers  is  33 
times  the  difference  of  the  two  digits,  and  the  difference  of 
the  squares  of  the  numbers  is  4752.     Find  the  numbers. 

16.  The  sum  of  the  numerator  and  denominator  of  a  cer- 
tain fraction  is  5 ;  and  if  the  numerator  and  denominator 
are  each  increased  by  3,  the  value  of  the  fraction  will  be 
increased  by  £.     Find  the  fraction. 

17.  The  fore  wheel  of  a  carriage  turns  in.  a  mile  132 
times  more  than  the  hind  wheel ;  but  if  the  circumferences 
were  each  increased  by  2  feet,  it  would  turn  only  88  times 
more.     Find  the  circumference  of  each. 


CHAPTER  XXL 
RATIO,  PROPORTION,  AND  VARIATION. 

316.  The  relative  magnitude  of  two  numbers  is  called 
their  ratio,  when  expressed  by  the  fraction  which  the  first 
is  of  the  second. 

Thus,  the  ratio  of  6  to  3  is  indicated  by  the  fraction  f ,  which  is 
sometimes  written  6  : 3. 

317.  The  first  term  of  a  ratio  is  called  the  antecedent, 
and  the  second  term  the  consequent.  When  the  antecedent 
is  equal  to  the  consequent,  the  ratio  is  called  a  ratio  of 
equality ;  when  the  antecedent  is  greater  than  the  conse- 
quent, the  ratio  is  called  a  ratio  of  greater  inequality  ;  when 
less,  a  ratio  of  less  inequality. 

318.  When  the  antecedent  and  consequent  are  inter- 
changed, the  resulting  ratio  is  called  the  inverse  of  the 
given  ratio. 

Thus,  the  ratio  3  :  6  is  the  inverse  of  the  ratio  6  : 3. 

319.  The  ratio  of  two  quantities  that  can  be  expressed 
in  integers  in  terms  of  a  common  unit  is  equal  to  the  ratio 
of  the  two  numbers  by  which  they  are  expressed. 

Thus,  the  ratio  of  $9  to  $11  is  equal  to  the  ratio  of  9  :  11 ;  and  the 
ratio  of  a  line  2|  inches  long  to  a  line  3|  inches  long,  when  both  are 
expressed  in  terms  of  a  unit  T2f  of  an  inch  long,  is  equal  to  the  ratio 
of  32  :  45. 

320.  Two  quantities  different  in  kind  can  have  no  ratio, 
for  then  one  cannot  be  a  fraction  of  the  other. 


304  RATIO. 

321.  Two  quantities  that  can  be  expressed  in  integers 
in  terms  of  a  common  unit  are  said  to  be  commensurable. 
The  common  unit  is  called  a  common  measure,  and  each 
quantity  is  called  a  multiple  of  this  common  measure. 

Thus,  a  common  measure  of  2|  feet  and  3|  feet  is  \  of  a  foot,  which 
is  contained  15  times  in  2£  feet,  and  22  times  in  8f  feet.  Hence,  2\ 
feet  and  3|  feet  are  multiples  of  J  of  a  foot,  2-J-  feet  being  obtained  by 
taking  £  of  a  foot  15  times,  and  3|  feet  by  taking  i  of  a  foot  22  times. 

322.  When  two  quantities  are  incommensurable,  that  is, 
have  no  common  unit  in  terms  of  which  both  quantities 
can  be  expressed  in  integers,  it  is  impossible  to  find  a  frac- 
tion that  will  indicate  the  exact  value  of  the  ratio  of  the 
given  quantities.  It  is  possible,  however,  by  taking  the 
unit  sufficiently  small,  to  find  a  fraction  that  shall  differ 
from  the  true  value  of  the  ratio  by  as  little  as  we  please. 

Thus,  if  a  and  &  denote  the  diagonal  and  side  of  a  square, 

Now  V2  =  1.41421356 ,  a  value  greater  than  1.414213,  but  less 

than  1.414214. 

If,  then,  a  millionth  part  of  b  is  taken  as  the  unit,  the  value  of  the 

ratio  -  lies  between  }ftfl|f|g  and  $$$$tBi  and  therefore  differs  from 

either  of  these  fractions  by  less  than  100^000. 

By  carrying  the  decimal  further,  a  fraction  may  be  found  that  will 
differ  from  the  true  value  of  the  ratio  by  less  than  a  billionth,  tril- 
lionth,  or  any  other  assigned  value  whatever. 

323.  Expressed  generally,  when  a  and  b  are  incommen- 
surable, and  b  is  divided  into  any  integral  number  (n)  of 
equal  parts,  if  one  of  these  parts  is  contained  in  a  more 
than  m  times,  but  less  than  m  +  1  times,  then 

a      m  ,  m  + 1 

T>->but< ; 

b      n  n 

that  is,  the  value  of  -  lies  between  —  and • 

'  b  n  n 


RATIO.  305 

The  error,  therefore,  in  taking  either  of  these  values  for 

-  is  less  than  the  difference  between  —  and ;  that  is, 

b  n  n      ' 

less  than  -•     But  by  increasing  n  indefinitely,  -  can  be 

made  to  decrease  indefinitely,  and  to  become  less  than  any 
assigned  value,  however  small,  though  it  cannot  be  made 
absolutely  equal  to  zero. 

324.    The  ratio  between  two  incommensurable  quantities 
is  called  an  incommensurable  ratio. 


325.  Theorem.  Two  incommensurable  ratios  are  equal 
if,  when  the  unit  of  measure  is  indefinitely  diminished,  their 
approximate  values  constantly  remain  equal. 

Let  a  :  b  and  a' :  V  be  two  incommensurable  ratios  whose 
true  values   lie   between   the   approximate   values   —   and 

>  when  the  unit  of  measure   is   indefinitely  dimin- 
ished.    Then  they  cannot  differ  by  so  much  as  -  • 

Now  the  difference  (if  any)  between  the  fixed  values  a :  b 
and  a' :  b'  is  a  fixed  value.     Let  d  denote  this  fixed  value. 

1 


Then,  d< 

n 

But  if  d  has  any  value,   however  small,  —  >   which  by 

hypothesis  can  be  made  less  than  any  value,  however  small, 
can  be  made  less  than  d. 


Therefore,  d  cannot  have  any  value ;  that  is,  d  =  0,  and 
ere  is  no  difference 
therefore,  a  :  b  =  a' :  b\ 


there  is  no  difference  between  the  ratios  a:b  and  a'lb'; 


306  RATIO. 

326.    A  ratio  will  not  be  altered  if  both  its  terms  are 
multiplied  by  the  same  positive  number. 

For  the  ratio  a :  b  is  represented  by  -  ■»  the  ratio  ma :  mb  is  repre- 
sented by  — -  :  and  since  — -  =  - '  therefore,  ma  :  mb  =  a  :  b. 
J  mb'  mb      b  ' 


327.  A  ratio  will  be  altered  if  its  terms  are  multiplied 
by  different  positive  numbers ;  and  will  be  increased  or 
diminished  according  as  the  multiplier  of  the  antecedent  is 
greater  than  or  less  than,  that  of  the  consequent. 

For  ma :  nb  >  or  <  a  :  b 

,.  ma .  a  /      na\ 

according  as  ri>0T<b{=nb} 

according  as  ma  >  or  <  na, 

according  as  m   >  or  <  n. 

328.  A  ratio  of  greater  inequality  will  be  diminished, 
and  a  ratio  of  less  inequality  increased,  by  adding  the  same 
positive  number  to  both  its  terms. 


For 

a  + 

x  :  &  +  x  >  or  <C.a:b. 

according  as 

a  +  x  ^             a 
6  +  x>°r<6' 

according  as 

ab  +  bx  >  or  <  ab  +  ax, 

according  as 

bx  >  or  <  ax, 

according  as 

6  >  or  <  a. 

329.  A  ratio  of  greater  inequality  will  be  increased,  and 
a  ratio  of  less  inequality  diminished,  by  subtracting  the  same 
positive  number  from  both  its  terms. 

For  a  —  x:b  —  x  >  or  <  a  :  &, 

_.  a  — x^    .a 

according  as         ,  _  >  or  <  - » 

according  as       ab  —  bx  >  or  <  ab  —  ax, 
according  as  ax  >  or  <  6x, 

according  as  a  >•  or  <  6. 


RA  TIO.  307 

330.  Katios  are  compounded  by  taking  the  product  of  the 
fractions  that  represent  them. 

Thus,  the  ratio  compounded  of  a  :  b  and  c  :  d  is  found  by  taking  the 

,  ,  a       ,c      ac 

product  of  -  and  -  =  ?-.  • 
b         a      bd 

The  ratio  compounded  of  a  :  b  and  a  ibis  the  duplicate  ratio  a2  :  62, 

and  the  ratio  compounded  of  a  :  6,  a  :  b,  and  a  :  6  is  the  triplicate  ratio 

a3  :  &8. 

331.  Ratios  are  compared  by  comparing  the  fractions  that 
represent  them. 

Thus,  a  :  b  >  or  <  c  :  d 

a.  e 


according  as  -  >  or  <  - » 


ad  ^bc 

M>0I<bd 
according  as  ad  >  or  <  be. 


according  as  —  >  or  <  ■ 


Exercise  123. 

1.  Write  the  ratio  compounded  of  3 :  5  and  8  :  7.  Which 
of  these  ratios  is  increased,  and  which  is  diminished  by  the 
composition  ? 

2.  Compound  the  duplicate  ratio  of  4  :  15  with  the  tripli- 
cate of  5 : 2. 

3.  Show  that  a  duplicate  ratio  is  greater  or  less  than  its 
simple  ratio  according  as  it  is  a  ratio  of  greater  or  less 
inequality. 

4.  Arrange  in  order  of  magnitude  the  ratios  3:4;  23  :  25 ; 
10  :  11 ;  and  15  :  16. 

5.  Arrange  in  order  of  magnitude 

a  -f  b  :  a  —  b  and  a2  -f  b2  :  a2  —  b2,  if  a  >  b. 
Find  the  ratio  compounded  of: 

6.  3:5;  10:21;  14:15.       7.    7:9;  102:105;  15:17. 


BATIO. 


a2  +  ax  -\-  x2  ^  a2  —  ax  -\-  x* 

8.  -= - — ■ r  and  ■ 

a9  —  azx  +  aar  —  x6  a  +  x 

n    ic2-9cc  +  20        ,  x2-13x  +  42 

9.   = - and  = = 

ar  —  o  x  ar  —  ox 

10.  a  +  bia-b;  a2  +  b2  :  (a  +  b)2-,   (a2-  b2)2 :  a4  -  £4. 

11.  Two  numbers  are  in  the  ratio  2  :  3,  and  if  9  is  added 
to  each,  they  are  in  the  ratio  3  :  4.     Find  the  numbers. 

Let  2  x  and  Sx  represent  the  numbers. 

12.  Show  that  the  ratio  a  :  b  is  the  duplicate  of  the  ratio 
a  +  c  :  b  4-  c,  if  c2  =  ab. 

]  3.  Two  numbers  are  in  the  ratio  3  :  4.  Their  sum  is  to 
the  sum  of  their  squares  as  7  :  50.     Find  the  numbers. 

14.  If  five  gold  and  four  silver  coins  are  worth  as  much 
as  three  gold  and  twelve  silver  coins,  find  the  ratio  of  the 
value  of  a  gold  coin  to  that  of  a.  silver  coin. 

15.  If  eight  gold  and  nine  silver  coins  are  worth  as  much 
as  six  gold  and  nineteen  silver  coins,  find  the  ratio  of  the 
value  of  a  silver  coin  to  that  of  a  gold  coin. 

16.  There  are  two  roads  from  A  to  B,  one  of  them  14 
miles  longer  than  the  other ;  and  two  roads  from  B  to  C, 
one  of  them  8  miles  longer  than  the  other.  The  distance 
from  A  to  B  is  to  the  distance  from  B  to  C,  by  the  shorter 
roads,  as  1  to  2 ;  by  the  longer  roads,  as  2  to  3.  Find  the 
distances. 

17.  What  must  be  added  to  each  of  the  terms  of  the  ratio 
m  :  n,  that  it  may  become  equal  to  the  ratio  p  :  q? 

18.  A  rectangular  field  contains  5270  acres,  and  its  length 
is  to  its  breadth  in  the  ratio  of  31 :  17.     Find  its  dimensions. 


PROPORTION.  309 

Proportion. 

332.  An  equation  consisting  of  two  equal  ratios  is  called 
a  proportion ;  and  the  terms  of  the  ratios  are  called  propor- 
tionals. 

333.  The  algebraic  test  of  a  proportion  is  that  the  two 
fractions  which  represent  the  ratios  of  the  quantities  com- 
pared shall  be  equal. 

Thus,  the  ratio  a :  b  is  equal  to  the  ratio  c  :  d  if  the  fraction  that 
represents  the  ratio  a :  b  is  equal  to  the  fraction  that  represents  the 
ratio  c  :  d.  Then  the  four  quantities,  a,  6,  c,  d,  are  called  proportionals, 
or  are  said  to  be  in  proportion. 

334.  If  the  ratios  a  :  b  and  c  :  d  form  a  proportion,  the 
proportion  is  written 

a  :  b  =  c  :  d 
(read  the  ratio  of  a  to  b  is  equal  to  the  ratio  of  c  to  d), 
or  a  :  b  : :  c  :  d 

(read  a  is  to  b  in  the  same  ratio  as  c  is  to  d). 

The  first  and  last  terms,  a  and  d,  are  called  the  extremes. 

The  two  middle  terms,  b  and  c,  are  called  the  means. 

335.  In  the  proportion  a  :  b  =  c  :  d ;  d  is  called  a  fourth 
proportional  to  a,  b,  and  c. 

In  the  proportion  a  :  b  =  b  :  c ;  c  is  called  a  £AiVd  propor- 
tional to  a  and  J. 

In  the  proportion  a  :  b  =  b  :  c ;  5  is  called  a  meaw  propor- 
tional between  a  and  c. 

336.  A  continued  proportion  is  a  series  of  equal  ratios  in 
which  each  consequent  is  the  same  as  the  next  antecedent. 

Thus,  a:&  =  6:c  =  c:d  =  d:e  =  e:/isa  continued  proportion. 


310  PROPORTION. 

337.  When  four  quantities  are  in  proportion,  the  product 
of  the  extremes  is  equal  to  the  product  of  the  means. 

For,  if  a  :  b  =  c  :  d, 

^  a  -  c . 

then,  l-- 

Multiply  by  bd,  ad  =  be. 

The  equation  ad  =  be  gives  a  =  —  >  b  =  — ;    so  that  an 

a  c 

extreme  may  be  found  by  dividing  the  product  of  the  means 
by  the  other  extreme ;  and  a  mean  may  be  found  by  divid- 
ing the  product  of  the  extremes  by  the  other  mean. 

Note.  By  the  product  of  two  quantities  we  mean  the  product 
of  the  two  numbers  that  represent  them  when  the  quantities  are 
expressed  in  a  common  unit. 


338.  If  the  product  of  two  quantities  is  equal  to  the  prod* 
net  of  two  others,  either  two  may  be  made  the  extremes  of 
a  proportion  and  the  other  two  the  means* 

For,  if 
then,  divide  by  bd, 


or 


339.     Transformations  of  a  Proportion.     If  four  quantities, 
a,  b,  c,  d,  are  in  proportion,  they  will  be  in  proportion  by : 

I.   Inversion ;  that  is,  b  will  be  to  a  as  d  is  to  c. 
For,  if  a:b  =  c:d, 

±v  a      c 


ad  — 

be, 

ad 
bd~ 

be 
bd! 

a. 
b~ 

c 
3" 

a\b  — 

■  c:d. 

PROPORTION.  311 

and  the  reciprocals  of  these  fractions  are  equal ; 

that  is,  -  =  -• 

a      c 

.'.  b:a  =  d  :  c. 
II.   Composition ;  that  is,  a  +  b  will  be  to  b  as  c  +  d  is  to  d. 
For,  if  a:b  =  c:df 


then, 


a  _c 
b~  d 


and  2  +  1  =  ^  +  1* 


or 


then, 


a  +  b  _  £_+d 
6      ~~_ d 
.'.  a  +  b:b  =  c  +  did. 

III.   Division ;  that  is,  <x  —  &  will  be  to  b  as  c  —  d  is  to  d. 

For,  if  a:b  =  c:d, 

a  _c 
b~  d' 


and  F"1^-1' 

a  —  b      c  ~  d 


or 


'.a  —  b  :b  =  c  —  did. 


IY.   Composition  and  Division ;   that  is,  a  +  b  will  be  to 
a  —  b  as  c  +  d  is  to  c  —  d. 

For,  from  II,  a  +  b      C  +  d 

and  from  III, 

Divide, 

a  —  b      c  —  d 

o\  a  +  b  :a  —  b  =  c  +  d:  c  —  d. 


b 

d 

a 

-b 

c  —  d 

b 

d 

a 

+  b 

c  +  d 

SV2 

PROPORTION. 

V.   Alternation; 

that  is,  a  will  be  to  c  as  b  is  to 

For,  if 

a  :  b  =  e  :  d, 

then, 

a       e 
b~d' 

Multiply  by  -> 

ab       be 
be       cd 

or 

a       b 

c       d 

.  \  a:  e  =  b  :d. 

Note.  In  order  for  four  quantities,  a,  6,  c,  d,  to  be  in  proportion, 
a  and  b  must  be  of  the  same  kind  and  c  and  d  of  the  same  kind ;  but 
c  and  d  need  not  necessarily  be  of  the  same  kind  as  a  and  6.  In 
applying  alternation,  however,  all  four  quantities  must  be  of  the  same 
kind. 

340.  In  a  Series  of  Equal  Ratios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 


-n      •*  a      e      e       g 

r  may  be  put  for  each  of  these  ratios. 

mi  a  c  e  a 

Then,  -  =  r,5  =  r,-  =  r,|  =  n 

.*.  a  —  br,  c  =  dr,  e  =fr,  g  =  hr. 
.'.  a  +  c  +  e  +  gr  =  (b  +  d  +f+  h)n 

a  +  c-\-  e  +  g  _      _  a 
'''b  +  d+f  +  h~r~V 
.'.  a  +  c+e  +  gib  +  d  +/+  A  =  a  :  &. 

In  like  manner  it  may  be  shown  that 

ma  +  nc  +j>e-h  qg  :mb  +  nd  +  pf  +  qhs=a:b. 


PROPORTION.  313 

341.  A  mean  proportional  between  two  quantities  is  equal 
to  the  square  root  of  their  product. 

For,  if  a:b  =  b:c, 

xv  a      b 

then,  T=~' 

'  b      o 

Clear  of  fractions,  b2  =  ac. 

Extract  the  square  root,   b  =  Vac. 

342.  The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 

For,  if  a:b  =  c.di 

e:f=g:h, 
k:l  =  m:n, 
a       c     e      g    k      m 

Take  the  product  of  the  left  members,  and  also  of  the 
right  members  of  these  equations, 
aek  _  cgm 
bfl       dhn 
.*.  aek  :  bfl  =  cgm  :  dhn. 

343.  Like  powers,  or  like  roots,  of  the  terms  of  a  propor- 
tion are  in  proportion. 

For,  if  a :  b  =  c  :  d, 


then, 

a 
b~ 

e 
*3r 

Raise  both  sides  to  the  nth. 

power, 

an 
bn~~ 

cn 
~  d» 

,»0  an  :  bn  =  cn  :  d» 


314  PROPORTION. 


Extract  the  nth  root,       -j  =  ■— • 

1111 
.*.  an  :  bn  =  cn  :  dn. 


344.  The  laws  that  have  been  established  for  ratios 
should  be  remembered  when  ratios  are  expressed  in  frac- 
tional form. 

,         .      x2  +  x  +  1      x2  -  x  -f  2 
1.    Solve  -j r  =  -j— -• 

By  composition  and  division, 

2  x2  2  x2 


2  (x  +  1)       -  2  (x  -  2) 

This  equation  is  satisfied  when  x  =  0.     For  any  other  value  of  x, 
we  may  divide  by  x2. 

We  then  have  — j-r  = » 

x  +  1      2  — a; 

and  therefore,  x  =  ^. 

2.    If  a  :  b  =  c  :  <£,  show  that 

a2  +  ab:b2-ab  =  c2  +  cd:d2-  cd. 


If 

o_  c 
b~d* 

then, 

a  +  b      c  +  d 
a  —  b      c—d 

(§  339,  IV) 

and 

a          c 
-b      -d 

"  — 6      a  —  b      —  d      c  —  d* 

(§342) 

that  is, 

a2  +  a&_c2  +  cd^ 
V>  —  ab      d*  —  cd 

or 

a2  +  ab  i  ft2  -  a&  =  c2  +  cd :  d2  -  cd„ 

PROPORTION.  315 

Exercise  124. 

1.  Find  a  third  proportional  to  21  and  28. 

2.  Find  a  mean  proportional  between  6  and  24 

3.  Find  a  fourth  proportional  to  3,  5,  and  42. 

4.  Find  a;  if  5  +  x  :  11  -  x  =  3  :  5. 

If  a  :  b  =  c  :  <7,  show  that : 

5.  ac:bd  =  c2:d2.  7.    a2  -  b2  :  c2  -  d2  =  a2 :  c*. 

6.  ab:cd  =  a2:c2.  8.    2a  +  b  :2c  +  d  =  b  ;d. 
9.    5a  —  b  :  5c  —  d  =  a:  c. 

10.  a-3£:a  +  36  =  c-3d:c  +  3d. 

11.  a2  +  a&  +  62:a2-aM-&2  =  c2  +  cd  +  d2:c2-cd  +  da. 

Find  x  in  the  proportion  : 

12.  45:68  =  90:z.  14.   x  :  1\  =  If  :  If . 

13.  6:3  =  a:  7.  15.   3  :  ic  =  7  :  42. 

16.  Find  two  numbers  in  the  ratio  2  :  3,  the  sum  of  whose 
squares  is  325. 

17.  Find  two  numbers  in  the  ratio  5 : 3,  thd  difference 
of  whose  squares  is  400. 

18.  Find  three  numbers  which  are  to  each  other  as 
2:3:5,  such  that  half  the  sum  of  the  greatest  and  least 
exceeds  the  other  by  25. 

19.  A  and  B  trade  with  different  sums.  A  gains  $200 
and  B  loses  $50  and  now  A's  stock  :  B's  : :  2  :  £.  But,  if  A 
had  gained  $100  and  B  lost  $85,  their  stocks  would  have 
been  as  15  :  3£,     Find  the  original  stock  of  each. 


316  VARIA  TION. 

20.  Find  a;  if  6x  -  a  :  ±x  -  b  =  3x  +  b  :2x  +  a. 

21.  Find  x  and  y  from  the  proportions 

x  :y  =  a;  -f  y  >  42 ;  x  :y  =  x  —  y  :6. 

22.  Find  <r  and  y  from  the  proportions 

2x  +  y:y  =  3y:2y  —  x', 
2x  +  l:2x  +  6  =  y:y  +  2. 

a+b+c+d     a—b+c—d  a      e 

23.  If  — — ■ -  = 1 —>  show  that  -  =  -=• 

a  +  b  —  c  —  d     a  —  b  —  c  +  d  b     d 

Variation. 

345.  One  quantity  is  said  to  vary  as  another,  when  tho 
two  quantities  are  so  related  that  the  ratio  of  any  two  values 
of  the  one  is  equal  to  the  ratio  of  the  corresponding  values 
of  the  other. 

Thus,  if  it  is  said  that  the  weight  of  water  varies  as  its  volume,  the 
meaning  is,  that  one  gallon  of  water  is  to  any  number  of  gallons  as  the 
weight  of  one  gallon  is  to  the  weight  of  the  given  number  of  gallons. 

346.  Two  quantities  may  be  so  related  that  when  a  value 
of  one  is  given,  the  corresponding  value  of  the  other  can  be 
found.  In  this  case  one  quantity  is  said  to  be  a  function 
of  the  other ;  that  is,  one  quantity  depends  upon  the  other 
for  its  value.  Thus,  if  the  rate  at  which  a  man  walks  is 
known,  the  distance  he  walks  can  be  found  when  the  time 
is  given ;  the  distance  is  in  this  case  a  funetion  of  the  time. 

347.  There  is  an  unlimited  number  of  ways  in  which 
tw©  quantities  may  be  related.  We  shall  consider  in  this 
chapter  only  a  few  of  these  ways. 

348.  When  x  and  y  are  so  related  that  their  ratio  is 
constant }  y  is  said  to  vary  as  x  \  this  is  abbreviated  thus : 


VARIATION.  317 

y  oo  x.      The  sign  oo,  called  the  sign  of  variation,  is  read 
varies  as. 

Thus,  the  area  of  a  triangle  with  a  given  base  varies  as  its  altitude ; 
for,  if  the  altitude  is  changed  in  any  ratio,  the  area  will  be  changed  in 
the  same  ratio. 

In  this  case,  if  we  represent  the  constant  ratio  by  m, 

V 
y  :  x  =  m,  or  -  =  m ;  .*.  y  ==  mx. 
x 

Again,  if  y\  x'  and  y",  x"  are  two  sets  of  corresponding 
values  of  y  and  x, 

then,  y1 :  x'  =  y"  :  x", 

or  y':y"  =  x':x".  (§339,  V) 

349.  When  x  and  y  are  so  related  that  the  ratio  of  y  to  - 

*      x 

is  constant,  y  is  said  to  vary  inversely  as  x ;  this  is  written 

1 

y  go  -• 
7      x 

Thus,  the  time  required  to  do  a  certain  amount  of  work  varies 
inversely  as  the  number  of  workmen  employed ;  for,  if  the  number  of 
workmen  is  doubled,  halved,  or  changed  in  any  other  ratio,  the  time 
required  will  be  halved,  doubled,  or  changed  in  the  inverse  ratio. 

In  this  case,  y :—  =  m \  .*.  y  =  —  >  and  xy  =  m;    that  is, 
x  x 

the  product  xy  is  constant. 

1  1 

As  before,  y:-  =  2/"::_, 

Jj  JO 

x'y'  =  x"y", 
or  y' :  y"  =  x"  :  x'.  (§  338) 

350.  If  the  ratio  of  y :  xz  is  constant,  then  y  is  said  to 
vary  jointly  as  x  and  z. 

In  this  case,  y  =  mxz, 

and  y'  :y"  =  x'z' :  x"z". 


318  VARIATION. 


x 

351.    If  the  ratio  y  :  -  is  constant,  then  y  varies  directly 
z 


as  x  and  inversely  as  z. 

TYbX 

In  this  case,  y  =  — > 


and  ?/ :  y"  =  — r :  — r-  =  —  :  —  • 

J    9  z'       z"        z'   z" 

352.  Theorem  1.     Ifyaox,  and  xao  z,  then  y  oo  z. 

For  y  =  mx  and  x  =  nz. 

.'.  y  =  mnz\ 
since  mw  is  constant,  y  varies  as  z. 

353.  Theorem  2.     If  y  go  x,  and  zaox,  then  (y  ±  2)  oo  a?. 
For  1/  =  m#  and  z  =  wsc. 

.'.  y  ±  z  =  (m  ±  w)  a; ; 
since  m  +  nis  constant,  y  ±z  varies  as  x. 

354.  Theorem  3.  If  y  00  x  when  z  is  constant,  and  y  00  z 
when  x  is  constant,  then  y  00  xz  when  x  and  z  are  both 
variable. 

Let  x',  y',  z',  and  x",  y",  z"  be  corresponding  values  of  the 
variables. 

Let  x  change  from  x'  to  x",  z  remaining  constant,  and  let 
the  corresponding  value  of  y  be  Y. 

Then,  y':Y=x':x".  (1) 

Now  let  z  change  from  z'  to  z",  x  remaining  constant. 

Then,  Y:y"  =  *':z".  (2) 

Multiply  (1)  and  (2), 

y'Y:y"Y=x'z,:x"z",  (§342) 

or  y' :  y"  =  x'z' :  x"z", 

or  y' :  x'z'  =  y" :  x"z".  (§  339,  V) 

.*.  the  ratio  y  :  xz  is  constant,  and  y  varies  as  xz. 


VARIATION.  319 

In  like  manner  it  may  be  shown  that  if  y  varies  as  each 
of  any  number  of  quantities  x,  z,  u,  etc.,  when  the  rest  are 
unchanged,  then  when  they  all  change,  y  oo  xzu,  etc. 

Thus,  the  volume  of  a  rectangular  solid  varies  as  Jhe  length  when 
the  width  and  thickness  remain  constant;  as  the  width  when  the 
length  and  thickness  remain  constant;  as  the  thickness  when  the 
length  and  width  remain  constant;  but  as  the  product  of  length, 
breadth,  and  thickness  when  all  three  vary. 

1.  If  y  varies  inversely  as  x,  and  when  y  =  2  the  cor- 
responding value  of  x  is  36,  find  the  corresponding  value 

of  x  when  y  =  9. 

m 
Here  y  =  — »  or  m  —  xy. 

.-.  m  =  2  X  36  =  72. 

If  9  and  72  are  substituted  for  y  and  m,  respectively, 

72 
the  result  is  9  =  —  i  or  9  x  =  72. 

x 

.-.  x  =  S. 

2.  The  weight  of  a  sphere  of  given  material  varies  as 
its  volume,  and  its  volume  varies  as  the  cube  of  its  diam- 
eter. If  a  sphere  4  inches  in  diameter  weighs  20  pounds, 
find  the  weight  of  a  sphere  5  inches  in  diameter. 

Let  W  represent  the  weight, 

V  represent  the  volume, 
and  D  represent  the  diameter. 

Then,  W  oo  V  and  V  oo  D8. 

.-.  TTooD8.  (§352) 

Put  W  =  mD3  ; 

then,  since  20  and  4  are  corresponding  values  of  W  and  D, 
20  =  m  x  64. 
20  _  5 
•••m-64-16* 
.*.  IF  =  £  2* 
.-.  when  D  =  5,  W **  A  °*  125  =  39xV 


320  VARIATION. 

Exercise  125. 

1.  If  x  oo  y,  and  if  y  —  3  when  x  =  5,  find  x  when  y  is  5. 

2.  If  JF  varies  inversely  as  P,  and  W  is  4  when  P  is 
15,  find  JFvhenPis  12. 

3.  If  x  oo  y  and  y  oo  z,  show  that  xz  oo  y2. 

4.  If  x  oo  -  and  y  oo  -?  show  that  a;  oo  2. 

5.  If  x  varies  inversely  as  y2  —  1,  and  is  equal  to  24 
when  y  —  10,  find  x  when  y  =  5. 

6.  If  sc  varies  as  -  -+-  ->  and  is  equal  to  3  when  y  =  1 

y  ■;-* 

and  «  =  2,  show  that  xyz  —  2(g  +  z). 

7.  If  x  —  y  varies  inversely  as  z  +  ->  and  x  +  y  varies 

z 

inversely  as  z >  find  the  relation  between  x  and  z  if 

x  =  l}  y  =  3,  when  £  =  -  • 

8.  The  area  of  a  circle  varies  as  the  square  of  its  radius, 
and  the  area  of  a  circle  whose  radius  is  1  foot  is  3.1416 
square  feet.  Find  the  area  of  a  circle  whose  radius  is  20 
feet. 

9.  The  volume  of  a  sphere  varies  as  the  cube  of  its 
radius,  and  the  volume  of  a  sphere  whose  radius  is  1  foot  is 
4.1888  cubic  feet.  Find  the  volume  of  a  sphere  whose  radius 
is  2  feet. 

10.  If  a  sphere  of  given  material  3  inches  in  diameter 
weighs  24  pounds,  how  much  will  a  sphere  of  the  same 
material  weigh  if  its  diameter  is  5  inches  ? 


VARIA  TION.  321 

11.  The  velocity  of  a  falling  body  varies  as  the  time 
during  which  it  has  fallen  from  rest.  If  the  velocity  of  a 
falling  body  at  the  end  of  2  seconds  is  64  feet,  what  is  its 
velocity  at  the  end  of  8  seconds  ? 

12.  The  distance  a  body  falls  from  rest  varies  as  the 
square  of  the  time  it  is  falling.  If  a  body  falls  through 
144  feet  in  3  seconds,  how  far  will  it  fall  in  5  seconds  ? 

The  volume  of  a  right  circular  cone  varies  jointly  as  its 
height  and  the  square  of  the  radius  of  its  base. 

13.  Compare  the  volume  of  two  cones,  one  of  which  is 
twice  as  high  as  the  other,  but  with  one  half  its  diameter. 

If  the  volume  of  a  cone  7  feet  high  with  a  base  whose 
radius  is  3  feet  is  66  cubic  feet : 

14.  Find  the  volume  of  a  cone  9  feet  high  with,  a  base 
whose  radius  is  3  feet. 

15.  Find  the  volume  of  a  cone  7  feet  high  with  a  base 
whose  radius  is  4  feet. 

16.  Find  the  volume  of  a  cone  9  feet  high  with  a  base 
whose  radius  is  4  feet. 

17.  The  volume  of  a  sphere  varies  as  the  cube  of  its 
radius.  If  the  volume  is  179§  cubic  feet  when  the  radius 
is  3£  feet,  find  the  volume  when  the  radius  is  1  foot  6 
inches. 

18.  Find  the  radius  of  a  sphere  whose  volume  is  the  sum 
of  the  volumes  of  two  spheres  with  radii  3£  feet  and  6  feet, 
respectively. 

19.  The  distance  of  the  offing  at  sea  varies  as  the  square 
root  of  the  height  of  the  eye  above  the  sea  level,  and  the 
distance  is  3  miles  when  the  height  is  6  feet.  Find  the 
distance  when  the  height  is  24  feet. 


CHAPTER   XXTL 
PROGRESSIONS. 

355.  A  succession  of  numbers  that  proceed  according  to 
some  fixed  law  is  called  a  series ;  the  successive  numbers  are 
called  the  terms  of  the  series. 

A  series  that  ends  at  some  particular  term  is  a  finite  series ; 
a  series  that  continues  without  end  is  an  infinite  series. 

Arithmetical  Progression. 

356.  A  series  is  called  an  arithmetical  series  or  an  arith- 
metical progression  when  each  term  after  the  first  is  obtained 
by  adding  to  the  preceding  term  a  constant  difference. 

The  general  representative  of  such  a  series  is 

1st  2d  3d  4th 

a,         a  +  d ,         a-{-2d,         a  +  3  d , 


in  which  a  is  the  first  term  and  d  the  common  difference ; 
the  series  is  increasing  or  decreasing  according  as  d  is 
positive  or  negative, 

357.  The  nth  Term.  Since  each  succeeding  term  of  the 
series  is  obtained  by  adding  d  to  the  preceding  term,  the 
coefficient  of  d  is  always  one  less  than  the  number  of 
the  term,  so  that  the  nth  term  is  a  -f-  (n  —  1)  d. 

If  the  nth.  term  is  represented  by  I,  we  have 
I  =  a  -f*  (n  —  l)d. 


ARITHMETICAL  PROGRESSION.  323 

358.  Sum  of  the  Series.  If  I  denotes  the  nth.  term,  a  the 
first  term,  n  the  number  of  terms,  d  the  common  difference, 
and  s  the  sum  of  n  terms,  it  is  evident  that 

s=     a      +  (a  +  d)  +  (a  +  2d)  + +  (l-d)  +     I,      or 

s=     I       4.  (Z -.^-f-(7  _2<Z)  + +  (a  +  rf)+     a, 

S.2s=(a  +  l)  +  (a  +  l)-\-(a  +  l)     + +  (a+Q+(a  +  0 

=  n(a  +  l). 

359.  From  the  equations 

I,  l  =  a  +  (n-l)d, 


any  two  of  the  five  numbers  a,  d,  I,  n,  s  may  be  found  when 
the  other  three  are  given. 

1.  Find  the  thirteenth  term  of  an  arithmetical  progres- 
sion, if  the  first  term  is  3  and  the  common  difference  5. 

Here  a  =  3,  d  =  5,  n  =  13. 

From  I,  I  =  3  +  (13  —  1)  5  =  63. 

2.  Find  the  arithmetical  series,  if  the  tenth  term  is  31 
and  the  twentieth  term  61. 

From  I,  a  +  19  d  =  61 

and  a+    9d  =  31 

Subtract  10  d  =  30 

Whence  d  =  3. 

Therefore,  a  =  4. 

Therefore,  the  series  is  4,  7,  10 


324  ARITHMETICAL  PROGRESSION. 

3.  Find  the  sum  of  ten  terms  of  the  series  2,  5,  8,  11 

Here  a  =  2,  d  =  3,  n  =  10. 

From  I,  I  =  2  +  27  =  29. 

Substitute  in  n,  s  =  ~  (2  +  29)  =  155. 

So 

The  sum  of  ten  terms  is  155. 

4.  The  first  term  of  an  arithmetical  series  is  3,  the  last 
term  31,  and  the  sum  of  the  series  136.     Find  the  series. 


From  I, 

81  =  3  +  (n  -  1 

From  II, 

136  =  |  (3 +  31) 

From  (2), 

n  =  8. 

Substitute 

in 

(1), 

d  =  4. 

The  series  is  3,  7,  11,  15,  19,  23,  27,  31. 

5.   How  many  terms  of  the  series  5,  9,  13 must  1 

taken  in  order  that  their  sum  may  be  275  ? 

From  I,  J  =  5+(n-l)4. 

.-.  Z  =  4n  +  1.  (1) 

FromH,  275  =  |  (5  +  0-  (2) 

Substitute  in  (2)  the  value  of  I  found  in  (1), 
275  =  |(4n  +  6), 
or  2  n2  +  3  n  =  275. 

Complete  the  square, 

16n2  +  (  ) +  9  =  2209. 
Extract  the  root,  4  n  +  3  =  ±  47. 

Therefore,  n  =  11,  or  —  12£. 

We  use  only  the  positive  result. 
Therefore,  11  terms  must  be  taken. 


ARITHMETICAL  PROGRESSION. 


325 


6.    Find  n  when  d,  lf  s  are  given. 
From  I, 
From  II, 


a  =  I  —  (n  —  l)<L 
2  s  — In 


Therefore, 


l-(n  —  l)d 


n 
2  s  — In 


In  —  dn2  +  dn  —  2s  —  ln. 
dn2  —  (2l  +  d)n-  —2s. 
Complete  the  square, 

4d2n2  -  (  )  +  (2 1  +  eZ)2  =  (21  +  d)*  -  Sds. 
Extract  the  root, 

2dn—  {21  +  d)  =  ±V(2J  +  d)2  -  8ds. 


.'.  n 


21  +  d  ±  V(2  I  +  d)2  -  sTs 


2d 


360.  The  arithmetical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
an  arithmetical  series. 

If  a  and  b  represent  two  numbers,  and  A  their  arithmet- 
ical mean,  then  a,  A,  b  are  in  arithmetical  progression. 

.*.  A  —  a  =  df         and  b  —  A  =  d. 

.'.  A  —  a  =  b  —  A. 

•  -  A  =  — r 


361.   Sometimes  it  is  required  to  insert  several  afithmet- 
teal  means  between  two  numbers. 

Insert  six  arithmetical  means  between  3  and  17. 
Here  the  whole  number  of  terms  is  eight ;  therefore,  by  I, 
17  =  3  +  7tf. 
.\  d  =  2. 

The  series  is  3,  [5,  7,  9,  11,  13,  15,]  17,  the  terms  in 
brackets  being  the  means  required. 

Note.    The  student  should  work  out  all  the  problems  on  the  follow- 
page,  using  the  formulas  I  and  H. 


326 


ARITHME TICAL  PROGRESSION. 


No. 

Given. 

Required. 

Formulas. 

1 
2 
3 
4 

adn 
ads 
an  s 
d  n  s 

1 

l  =  a+(n-l)d. 

l  =  —±d±  V2ds-f  (a  — id)*. 

I  = a. 

8  +.(n-l)d. 
n            2 

5 
6 

7 
8 

adn 
a  d  I 
a  n  I 
d  n  I 

s 

s  =  in[2a  +  (n  —  l)d]. 

l  +  a  .  Z2-a2 
S=     2     +     2d     ' 

s=(l  +  a)^ 

s  =  in[2l—(n  —  l)d]. 

9 
10 
11 
12 

d  n  I 
dn  s 
d  Is 
n  I  s 

a 

a  =  I  —  (n  —  1)  d. 
s       (n  —  1)  d 
a=n             2        * 

a  =  id±V(l  +  id)2-2ds. 

2s 

a  = 1. 

n 

13 
14 
15 
16 

a  n  I 
an  s 
a  I  s 
n  Is 

d 

,       I  —  a 
d=            • 
n  —  1 

d_2(s-an)i 
n(n  —  l) 
P  -  a2 
28  —  I  —  a 

M      2  (In-  s) 
n  (n  —  1) 

17 
18 
19 
20 

a  d  I 
ads 
a  Is 
dls 

n 

n  =  — -=—  +  1. 
a 

d  -  2  a  ±  V(2  a  —  d)2  +  8  ds 
n~                      2d 

2s 

n  =  — — • 
l  +  a 

2Z  +  d±V(2Z  +  d)2-8ds 
n  = 2d 

ARITHMETICAL  PROGRESSION.  327 

Exercise  126. 

1.  Find  the  tenth  term  of  9,  13,  17 

2.  Find  the  thirteenth  term  of  —  3,  —  1,  -f-  1  

3.  Find  the  ninth  term  of  —  5,  —  8,  —  11 

4.  Find  the  eighth  term  of  a,  a  +  3  b,  a  -f  6  b 

6.  Find  the  fifteenth  term  of  1,  f ,  f 

6.  Find  the  fourteenth  term  of  -  44,  -  40,  —  36 

7.  The  first  term  of  an  arithmetical  series  is  3,  the  thir- 
teenth term  is  55.     Find  the  common  difference. 

8.  Find   the  arithmetical  mean  between  —  5  and  17 ; 
between  a2  +  ab  +  b2  and  a2  -  ab  +  b2. 

9.  Insert  three  arithmetical  means  between  1  and  19; 
and  four  between  —  4  and  17. 

10.  The  first  term  of  a  series  is  2,  and  the  common  differ- 
ence £.     What  term  will  be  12  ? 

1 1 .  The  seventh  term  of  a  series,  whose  common  difference 
is  3,  is  11.     Find  the  first  term. 

Find  the  sum  of : 

12.  5  +  8  +  11  + to  ten  terms. 

13.  —  4  —  1  -f  2  + to  seven  terms. 

14.  a  +  4«-f  7a+ to  n  terms. 

15.  §  +  ^3-  +  ^  + to  twenty-one  terms. 

16.  1  -f  2§  +  4£  + to  twenty  terms. 

17.  The  sum  of  six  terms  of  an  arithmetical  series  is  27, 
and  the  first  term  is  1.     Find  the  series. 

18.  How   many  terms  of  the   series  —  5,  —  2,  + 1 

must  be  taken  that  their  sum  may  be  63  ? 

19.  The  first  term  of  an  arithmetical  series  is  12,  and  the 
sum  of  ten  terms  is  10.     Find  the  tenth  term. 


328  ARITHMETICAL  PROGRESSION. 

20.  When  a  train  arrives  at  the  top  of  a  long  slope,  the 
last  car  is  detached  and  begins  to  descend,  passing  over  3 
feet  in  the  first  second,  three  times  3  feet  in  the  second 
second,  five  times  3  feet  in  the  third  second,  etc.  At  the 
end  of  30  seconds  it  reaches  the  bottom  of  the  slope.  Find 
its  velocity  the  last  second. 

21.  Insert  eleven  arithmetical  means  between  1  and  12. 

22.  The  first  term  of  an  arithmetical  series  is  3,  and  the 
sum  of  six  terms  is  28.     What  term  will  be  9  ? 

23.  The  arithmetical  mean  between  two  numbers  is  10, 
and  the  mean  between  the  double  of  the  first  and  the 
triple  of  the  second  is  27.     Find  the  numbers. 

24.  The  first  term  of  an  arithmetical  progression  is  3, 
and  the  third  term  is  11.     Find  the  sum  of  seven  terms. 

25.  A  common  clock  strikes  the  hours  from  1  to  12. 
How  many  times  does  it  strike  every  24  hours  ? 

26.  The  Greenwich  clock  strikes  the  hours  from  1  to  24. 
How  many  times  does  it  strike  in  24  hours  ? 

27.  Find  three  numbers  in  arithmetical  progression  of 
which  the  sum  is  21,  and  the  sum  of  the  first  and  second 
j  of  the  sum  of  the  second  and  third. 

Let  x  —  y,  x,  and  x  +  y  stand  for  the  numbers. 

28.  The  sum  of  three  numbers  in  arithmetical  progres- 
sion is  33,  and  the  sum  of  their  squares  is  461.  Find  the 
numbers. 

29.  The  sum  of  four  numbers  in  arithmetical  progres- 
sion is  12,  and  the  sum  of  their  squares  is  116.  What  are 
the  numbers  ? 

Let  x  —  3  y,  x  —  y,  z  +  y,  and  x  +  3  y  stand  for  the  numbers. 


GEOMETRICAL  PROGRESSION.  329 

Geometrical  Progression. 

362.  A  series  is  called  a  geometrical  series  or  a  geometrical 
progression  when  each  succeeding  term  is  obtained  by  mul 
tiplying  the  preceding  term  by  a  constant  multiplier. 

The  general  representative  of  such  a  series  is 

a }  ar,  ar2,  arz,  arA , 

in  which  a  is  the  first  term  and  r  the  constant  multiplier,  or 
ratio. 

The  terms  increase  or  decrease  in  numerical  magnitude 
according  as  r  is  numerically  greater  than  or  numerically 
less  than  unity. 

363.  The  nth  Term.  Since  the  exponent  of  r  increases 
by  one  for  each  succeeding  term  after  the  first,  the  exponent 
is  always  one  less  than  the  number  of  the  term,  so  that 
the  nth  term  is  arn~x. 

If  the  nth  term  is  represented  by  I,  we  have 
I,  1  =  ar""1. 

364.  Sum  of  the  Series.  If  I  represents  the  nth.  term,  a  the 
first  term,  n  the  number  of  terms,  r  the  common  ratio,  and 
s  the  sum  of  n  terms,  then, 

s  =  a  +  ar  +  ar2  + +  ar""1.  (1) 

Multiply  by  r, 

rs  =  ar  +  ar2  +  ar8  +  •— •  +  ar""1  +  ar",     (2) 
Subtract  (1)  from  (2), 
rs  —  s  —  ar11  —  a, 
or        (r  —  1)  s  =  a  (r*  —  1). 

n,  .-..-'fr-1*- 

■  r  —  1 


330  GEOMETRICAL  PROGRESSION. 

365.  When  r  <  1,  formula  II  may  be  more  conveniently 
written 

a  (1  -  r") 
1  —  r 

366.  I  =  arn~\  (§  363) 
Multiply  by  r,             rl  =  arn. 

By  putting  rl  for  arn  in  formula  II,  we  have 

ttt  rl  —  a 

III,  s  = r- 

r  —  1 

When  t*  <  1,  formula  III  is  more  conveniently  written 

a  —  ri 

367.  From  the  two  formulas  I  and  II,  or  the  two  for- 
mulas I  and  III,  any  two  of  the  five  numbers  a,  r,  I,  nT  s 
may  be  found  when  the  other  three  are  given. 

1.  Find  the  tenth  term  of  a  geometrical  series  if  the 
first  term  is  3  and  the  ratio  2. 

Here  a  =  3,  r  =  2,  n  =  10. 

From  I,  I  =  3  X  29  =  3  X  512  =  1536. 

2.  Find  the  geometrical  series  if  the  third  term  is  20  and 
the  sixth  term  160. 

Let  a  =  the  first  term,  and  r  =  the  ratio. 

Then,  ar2  =  the  third  term,  and  ar5  =  the  sixth  term. 

_      ,        ar*      160 
Therefore,  -  =  —  - 

r«  =  8. 
. .  r  ~  2. 
Since  ar2  =  20,  a  =  20  +  4  =  6. 

The  series  is  5,  10,  20,  40 


3.   Find  the  sum  of  six  terms  of  the  series  3,  6,  12 
Here  a  =  3,  r  —  2,  n  =  6. 

1  r— 1  r— 1 


192 

=  3r»-i. 

381 

192  r  — 

3 

r-1 

r 

=  2. 

2»-i 

=  64. 

.-.  n 

=  7. 

GEOMETRICAL  PROGRESSION.  331 

4.   The  first  term  of  a  geometrical  series  is  3,  the  last 

term  192,  and  the  sum  of  the  series  381.     Find  the  number 

of  terms  and  the  ratio. 

From  I,  192  =  3r»-i.  (1) 

i  qo  r o 

From  III,  381  =     m  _  ,     .  (2) 

From  (2), 
Substitute  in  (1), 

The  series  is  3,  6,  12,  24,  48,  96,  192. 

368.  The  geometrical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
a  geometrical  series. 

If  a  and  b  denote  two  numbers,  and  G  their  geometrical 
mean,  then  a,  G,  b  are  in  geometrical  progression,  and  by 
the  definition  of  a  geometrical  series  (§  362), 

G  ,    b 

—  =  r,  and  —  =  r» 

a  G 

•   ~  =  A 
"a       G 

.'.  G=^fab. 

369.  Sometimes  it  is  required  to  insert  several  geomet- 
rical means  between  two  numbers. 

Insert  three  geometrical  means  between  3  and  48. 

Here  the  whole  number  of  terms  is  five ;  3  is  the  first  term  and  48 
the  fifth. 

By  I,  48  =  3r*. 

r*  =  16. 
.-.r=±2. 

The  series  is  3,    [     6,  12,       24],  48; 

or  3,    [-6,  12,  -24],  48. 

The  terms  in  brackets  are  the  means  required 

In  working  out  the  following  results,  the  student  shoald  make  use 
of  formulas  I,  II,  and  ILL 


332 


GEOMETRICAL  PROGRESSION. 


No. 

Given. 

Required. 

Formulas. 

1 

2 
3 
4 

am 
ar  s 
an  s 
r  n  s 

1 

1  =  ar1-1. 
l_a  +  (r-l)s_ 
r 

I  (s  —  O""1  —  a  (s  —  a)n~l  =  0. 
_  (r  -  1)  sr*-1 . 
rn  —  1 

5 
6 

7 

8 

ar  n 
a  r  I 

a  n  I 

ml 

8 

.  _  a  (r"  -  1) 
r  —  1. 
rl  —  a 
S=  r-l   ' 

n  —  1, —        n  —  1. — 

S  — : : 

n  —  1  .           n—  1 ,— 

Vi  -    v^ 

lr*-l 

8  = r  * 

yn  —  pa  —  1 

9 
10 
11 
12 

r  n  I 
r  n  s 
r  I  s 
n  I  s 

a 

_        I 
f-n  —  1 

(r-l)  s 
rn  —  1 

a  =  rl  —  (r  —  1)  s. 

a(s  —  a)"-1  -  J  (s  -  l)»-  J  =  0. 

13 
14 
15 
16 

a  n  I 
an  s 
a  I  s 
n  I  8 

r 

s     .  s  —  a     rt 

rn r  -J =  0. 

a           a 
s  —  a 

r" .  r"-1  H .  =  0. 

s  —  l               s  —  l 

GEOMETRICAL  PROGRESSION.  333 

Exercise  127. 

X.  Find  the  seventh  term  of  2,  6,  18 

2.  Find  the  sixth  term  of  3,  6,  12 

3.  Find  the  ninth  term  of  6,  3,  lj- 

4.  Find  the  eighth  term  of  1,  —  2,  4 

5.  Find  the  fifth  term  of  4  a,  —  6  ma2,  9  m2a8 

6.  Find  the  geometrical  mean  between  18  x*y  and  3<)xyzz. 

7.  Find  the  ratio  when  the  first  and  third  terms  are  5 
and  80,  respectively. 

8.  Insert  two  geometrical  means  between  8  and  125; 
and  three  between  14  and  224. 

9.  If  a  =  2  and  r  =  3,  which  term  will  be  equal  to  162  ? 

10.  The  fifth  term  of  a  geometrical  series  is  48,  and  the 
ratio  2.     Find  the  first  and  seventh  terms. 

Find  the  sum  of : 

11.  3  +  6  +  12  + to  eight  terms. 

12.  1  —  34-9  — to  seven  terms. 

13.  8  +  4  +  2  -f  to  ten  terms. 

14.  0.1  +  0.5  4-  2.5  4- to  seven  terms. 

15.  m  — 7  4-  rrz  — to  five  terms. 

4      16 

16.  The  population  of  a  city  increased  in  four  years  from 
10,000  to  14,641.     What  was  the  annual  rate  of  increase  ? 

17.  The  sum  of  four  numbers  in  geometrical  progression 
is  200,  and  the  first  term  is  5.     Find  the  ratio. 

18.  Find  the  sum  of  eight  terms  of  a  geometrical  series 
whose  last  term  is  1,  and  fifth  term  £. 


334  GEOMETRICAL  PROGRESSION. 

19.  In  an  odd  number  of  terms,  show  that  the  product  of 
the  first  and  last  is  equal  to  the  square  of  the  middle  term. 

20.  The  product  of  four  terms  of  a  geometrical  series  is 
4,  and  the  fourth  term  is  4.     Find  the  series. 

21.  If  from  a  line  one  third  is  cut  off,  then  one  third  of 
the  remainder,  and  so  on,  what  fraction  of  the  whole  will 
remain  when  this  has  been  done  five  times  ? 

22.  Of  three  numbers  in  geometrical  progression,  the 
sum  of  the  first  and  second  exceeds  the  third  by  3,  and  the 
sum  of  the  first  and  third  exceeds  the  second  by  21.  What 
are  the  numbers  ? 

23.  Find  two  numbers  whose  sum  is  3£  and  geometrical 
mean  l£. 

24.  The  sum  of  the  squares  of  two  numbers  exceeds  twice 
their  product  by  576 ;  the  arithmetical  mean  of  the  two 
numbers  exceeds  the  geometrical  mean  by  6.  Find  the 
numbers. 

25.  There  are  four  numbers  such  that  the  sum  of  the 
first  and  the  last  is  11,  and  the  sum  of  the  other  two  is  10. 
The  first  three  of  these  four  numbers  are  in  arithmetical 
progression,  and  the  last  three  are  in  geometrical  progres- 
sion.    Find  the  numbers. 

26.  Find  three  numbers  in  geometrical  progression  such 
that  their  sum  is  13  and  the  sum  of  their  squares  91. 

27.  The  difference  between  two  numbers  is  48,  and  the 
arithmetical  mean  exceeds  the  geometrical  mean  by  18. 
Find  the  numbers. 

28.  There  are  four  numbers  in  geometrical  progression, 
the  second  of  which  is  less  than  the  fourth  by  24,  and  the 
sum  of  the  extremes  is  to  the  sum  of  the  means  as  7  to  3. 
Find  the  numbers. 


GEOMETRICAL  PROGRESSION.  335 


Infinite  Geometrical  Series. 

370.  When  r  is  less  than  1,  the  successive  terms  of  a 
geometrical  series  become  numerically  smaller;  by  taking  n 
large  enough  we  can  make  the  nth.  term,  arn~1f  as  small  as 
we  please,  although  we  cannot  make  it  absolutely  zero. 

The  sum  of  n  terms,  -r— * —  (§  366),  may  be  written 

: - ;    this  sum  differs   from   : by   - : 

1  —  r       1  —  r\  1  —  r     J    1  —  r' 

by  taking  enough  terms  we  can  make  I,  and  consequently 
j  as  small  as  we  please.  Hence, may  be  con- 
sidered the  sum  of  an  infinite  number  of  terms  of  the  series. 

1.  Find  the  sum  of  1  -  j  +  J  -  J  -f 

Here  a  =  1,  and  r  =  —  £. 

_     a     _         1         _      1      _2 

2.  Find  the  value  of  0.2363636 

The  terms  after  the  first  form  a  decreasing  geometrical  series  in 
which  a  =  0.036,  and  r  =  0.01. 

_     a  0.036      _  0.036  _  36  _  2 

*'•  S      1  -  r  ~  1  -  0.01  ~  0.99  ~  990  ~  55*' 

2        2       22  +  4      13 
Therefore,  the  required  value  is   —  +  —  =     rrr     =  —  * 

10      55  110         55 


Exercise  128. 
Find  the  sum  of  the  following  infinite  series : 
1.    4  +  2  +  1  + 5.    2-li  +f- 

2-  £  +  £  +  !  + 6.   0.1  +  0.01  +  0.001  + 

3-  i  -  rV  +  &  - 7.    0.868686 

4-  1  -  *  +  A  - 8-    0.54444 


336  HABMONICAL  PROGRESSION. 

♦Harmonica!  Progression. 

371.  A  series  is  called  a  harmonical  series  or  a  harmonical 
progression  when  the  reciprocals  of  its  terms  form  an  arith- 
metical series. 

Hence,  the  general  representative  of  such  a  series  is 

1        1        1_  1 

a    a-\r  d    a  -{-2d         '    a  -{-  (n  —  1)  d 

372.  Questions  relating  to  harmonical  series  should  be 
solved  by  writing  the  reciprocals  of  its  terms  so  as  to  form 
an  arithmetical  series. 

373.  If  a  and  b  denote  two  numbers,  and  H  their  har- 
monical mean,  then,  by  the  definition  of  a  harmonical  series. 

H      a      b      H 

.  1.  _  1  4. 1  -  a  +  b- 
'  '  H      a       b         ab 

2ab 


\H  = 


a  +  b 


374.  Sometimes  it  is  required  to  insert  several  harmon- 
ical means  between  two  numbers. 

Let  it  be  required  to  insert  three  harmonical  means  be- 
tween 3  and  18. 
0 

Find  the  three  arithmetical  means  between  £  and  T\. 

These  are  found  to  be  |f,  ^f ,  fy ;  therefore,  the  harmonical  means 
*»**.**.¥;  or8H,6fc8. 

*  A  harmonical  series  is  so  called  because  musical  strings  of  uniform 
thickness  and  tension  produce  harmony  when  their  lengths  are  repre- 
sented by  the  reciprocals  of  the  natural  series  of  numbers ;  that  is,  by 
the  harmonical  series  1,  i,  £,  £,  §,  etc, 


HARMONICAL  PROGRESSION.  337 

Exercise  129. 

1.  Insert  four  harmonical  means  between  2  and  12. 

2.  Find  two  numbers  whose  difference  is  8  and  har- 
monical mean  If. 

3.  Find  the  seventh  term  of  the  harmonical  series  3, 

3M 

4.  Continue  to  two  terms  each  way  the  harmonical 
series  two  consecutive  terms  of  which  are  15,  16. 

5.  The  first  two  terms  of  a  harmonical  series  are  5  and 
6.     Which  term  will  equal  30  ? 

6.  The  fifth  and  ninth  terms  of  a  harmonical  series  are 
8  and  12.     Find  the  first  four  terms. 

7.  The  difference  between  the  arithmetical  and  har- 
monical means  between  two  numbers  is  If,  and  jone  of  the 
numbers  is  four  times  the  other.     Find  the  numbers. 

8.  Find  the  arithmetical,  geometrical,  and  harmonical 
means  between  two  numbers,  a  and  b ;  and  show  that  the 
geometrical  mean  is  a  mean  proportional  between  the 
arithmetical  and  harmonical  means.  Also,  arrange  these 
means  in  order  of  magnitude. 

9.  The  arithmetical  mean  between  two  numbers  exceeds 
the  geometrical  by  13,  and  the  geometrical  exceeds  the 
harmonical  by  12.     What  are  the  numbers  ? 

10.  The  sum  of  three  terms  of  a  harmonical  series  is  11, 
and  the  sum  of  their  squares  is  49.     Find  the  numbers. 

11.  When  a,  b,  c  are  in  harmonical"  progression,  show 
that  a  :  c  :  :  a  —  b  :  b  —  c. 


CHAPTER  XXIII. 
VARIABLES  AND  LIMITS. 

375.  Constants  and  Variables.  A  number  that,  under  the 
conditions  of  the  problem  into  which  it  enters,  may  be 
made  to  assume  any  one  of  an  unlimited  number  of  values 
is  called  a  variable. 

A  number  that,  under  the  conditions  of  the  problem  into 
which  it  enters,  has  a  fixed  value  is  called  a  constant. 

Variables  are  represented  by  x,  y,  z ;  constants  by  a,  b,  c, 
and  by  the  Arabic  numerals. 

376.  Limits.  When  the  value  of  a  variable,  measured  at 
a  series  of  definite  intervals,  can  by  continuing  the  series 
be  made  to  differ  from  a  given  constant  by  less  than  any 
assigned  quantity,  however  small,  but  cannot  be  made  abso- 
lutely equal  to  the  constant,  the  constant  is  called  the  limit 
of  the  variable,  and  the  variable  is  said  to  approach  indefi- 
nitely to  its  limit. 

Consider  the  repetend  0.33&.-.-,  which  may  be  written 

■$5  +  "dhy "+"  TtfW  + 

The  value  of  each  fraction  after  the  first  is  one  tenth  of 

the  preceding  fraction,  and  by  continuing  the  series  we 
shall  reach  a  fraction  less  than  any  assigned  value,  how- 
ever small ;  that  is,  the  values  of  the  successive  fractions 
approach  0  as  a  limit. 

The  sum  of  these  fractions  will  always  be  less  than  £ ; 
but  the  more  terms  we  take,  the  nearer  does  the  sum 
approach  ^  as  a  limit. 


VARIABLES  AND  LIMITS.  339 

Suppose  a  point  to  move  from  A  toward  B,  under  the  con- 
ditions that  the  first  A  mm-  m-  b 
second  it  shall  move  ~~  '  '  ' 
one  half  the  distance  from  A  to  B,  that  is,  to  31;  the  next 
second,  one  half  the  remaining  distance,  that  is,  to  M' ;  the 
next  second,  one  half  the  remaining  distance,  that  is,  to  M " ; 
and  so  on  indefinitely. 

Then  it  is  evident  that  the  moving  point  may  approach 
as  near  to  B  as  we  please,  but  will  never  arrive  at  B.  For, 
however  near  it  may  be  to  B  at  any  instant,  the  next  second  it 
will  pass  over  one  half  the  interval  still  remaining ;  it  must, 
therefore,  approach  nearer  to  B,  since  half  the  interval 
still  remaining  is  some  distance,  but  will  not  reach  B  since 
half  the  interval  still  remaining  is  not  the  whole  distance. 

Hence,  the  distance  from  A  to  the  moving  point  is  an 
increasing  variable,  which  indefinitely  approaches  the  con- 
stant AB  as  its  limit ;  and  the  distance  from  the  moving 
point  to  B  is  a  decreasing  variable,  which  indefinitely 
approaches  the  constant  zero  as  its  limit. 

If  the  length  of  AB  is  two  inches,  and  the  variable  is 
denoted  by  x,  and  the  difference  between  the  variable  and 
its  limit  by  y : 

after  one  second,        x  =  1,  y=X 

after  two  seconds,      x  =  1  +  \,  y  =  \ 

after  three  seconds,  x  =  \-\-\  +  \,  V  —  \ 
after  four  seconds,  «  =  l  +  £  +  i+£,  y '  =  \ 
and  so  on  indefinitely. 

Now  the  sum  of  the  series  1  +  \  4-  \  +  \>  etc.,  is  less  than 
2 ;  but  by  taking  a  great  number  of  terms,  the  sum  can  be 
made  to  differ  from  2  by  as  little  as  we  please.  Hence,  2  is 
the  limit  of  the  sum  of  the  series,  when  the  number  of  the 
terms  is  increased  indefinitely;  and  0  is  the  limit  of  the 
difference  between  this  variable  sum  and  2. 


340  VARIABLES  AND  LIMITS. 

377.  Test  for  a  Limit.  In  order  to  prove  that  a  variable 
approaches  a  constant  as  a  limit,  it  is  necessary  and  suffi- 
cient to  prove  that  the  difference  between  the  variable  and 
the  constant  can  be  made  as  near  to  0  as  we  please,  but 
cannot  be  made  absolutely  equal  to  0. 

A  variable  may  approach  a  constant  without  approaching  it  as  a 
limit.  Thus,  in  the  last  example  x  approaches  3,  but  not  as  a  limit ; 
for  3  —  x  cannot  be  made  as  near  to  0  as  we  please,  since  it  cannot 
be  made  less  than  1. 

378.  Infinites.  As  a  variable  changes  its  value,  it  may 
constantly  increase  in  numerical  value ;  if  the  variable  can 
become  numerically  greater  than  any  assigned  value,  how- 
ever great  this  assigned  value  may  be,  the  variable  is  said 
to  increase  without  limit,  or  to  increase  indefinitely. 

When  a  variable  is  conceived  to  have  a  value  greater 
than  any  assigned  value,  however  great,  the  variable  is  said 
to  become  infinite;  such  a  variable  is  called  an  infinite 
number,  or  simply  an  infinite,  and  is  denoted  by  oo. 

379.  Infinitesimals.  As  a  variable  changes  its  value,  it 
may  constantly  decrease  in  numerical  value;  if  the  vari- 
able can  become  numerically  less  than  any  assigned  value, 
however  small  this  assigned  value  may  be,  the  variable  is 
said  to  decrease  without  limit,  or  to  decrease  indefinitely. 

In  this  case  the  variable  approaches  0  as  a  limit. 

When  a  variable  which  approaches  0  as  a  limit  is  con- 
ceived to  have  a  value  less  than  any  assigned  value,  how- 
ever small  this  assigned  value  may  be,  the  variable  is  said 
to  become  infinitesimal ;  such  a  variable  is  called  an  infini- 
tesimal number,  or  simply  an  infinitesimal. 

380.  Infinites  and  infinitesimals  are  variables,  not  con- 
stants. There  is  no  idea  of  fixed  value  implied  in  either 
an  infinite  or  an  infinitesimal. 


VARIABLES  AND  LIMITS.  341 

381.  An  infinitesimal  is  not  0.  An  infinitesimal  is  a 
variable  arising  from  the  division  of  a  quantity  into  a  con- 
stantly increasing  number  of  parts  ;  0  is  a  constant  arising 
from'  taking  the  difference  of  two  equal  quantities. 

382.  Finite  Numbers.  A  number  which  cannot  become 
infinite  is  said  to  be  finite. 

Theorems  of  Infinites  and  Infinitesimals. 

383.  Theorem  1.  If  x  is  infinitesimal,  and  a  is  finite 
and  not  0,  then  ax  is  infinitesimal. 

For  ax  can  be  made  as  small  as  we  please  since  x  can  be 
made  as  small  as  we  please. 

384.  Theorem  2.  If  X  is  infinite,  and  a  is  finite  and 
not  0,  then  aX  is  infinite. 

For  aX  can  be  made  as  large  as  we  please  since  Xcan  be 
made  as  large  as  we  please. 

385.  Theorem  3.  Ifx  is  infinitesimal,  and  a  is  finite  and 
not  0,  then  —  is  infinite. 

For  -  can  be  made  as  large  as  we  please  since  x  can  be 

made  as  small  as  we  please. 
t 

386.  Theorem  4.     If  X  is  infinite,  and  a  is  finite  and 

not  0,  then  —  is  infinitesimal. 

For  —  can  be  made  as  small  as  we  please  since  X  can  be 

made  as  large  as  we  please. 

In  the  above  theorems  a  may  be  a  constant  or  a  variable ; 
the  only  restriction  on  the  value  of  a  is  that  it  shall  not 
become  either  infinite  or  0. 


342  VARIABLES  AND  LIMITS. 

387.  Abbreviated  Notation.     The  expression  -  cannot  be 

interpreted  literally,  since  we  cannot  divide  by  0 ;   neither 

can  —  =  0  be  interpreted  literally,  since  we  can  find  no 
oo 

number  such  that  the  quotient  obtained  by  dividing  a  by 

that  number  is  0. 

The  expression  -  =  oo  is  simply  an  abbreviated  way  of  writing : 

Lf-  =  X,  and  x  approaches  0  as  a  limit,  X  increases  without  limit, 
x 

The  expression  —  =  0  is  simply  an  abbreviated  way  of  writing : 
If  -=.  =  x,  and  X  increases  without  limit,  x  approaches  Oas  a  limti. 

388.  Approach  to  a  Limit.     A  variable  may  approach  its 
limit  in  one  of  three  ways : 

1.  The  variable  may  be  always  less  than  its  limit. 

2.  The  variable  may  be  always  greater  than  its  limit. 

3.  The  variable  may  be  sometimes  less  and  sometimes 
greater  than  its  limit. 


If  x  represent  the  sum  of  n  terms  of  the  series  1  +  i  +  i  +  \  + , 

x  is  always  less  than  its  limit  2. 

If  x  represent  the  sum  of  n  terms  of  the  series  3  —  ■$-  —  £■  —  \  — , 

x  is  always  greater  than  its  limit  2. 

If  x  represent  the  sum  of  n  terms  of  the  series  3  —  |  +  |  —  f+' , 

we  have  (§  364) 

As  n  is  indefinitely  increased,  x  evidently  approaches  2  as  a  limit. 
If  n  is  even,  x  is  less  than  2 ;  if  n  is  odd,  x  is  greater  than  2. 
Hence,  if  n  is  increased  by  taking  each  time  one  more  term,  x  will 
be  alternately  less  than  and  greater  than  2.     If,  for  example, 

n=  2,         3,         4,         5,  6,  7, 

x  =  li,       2i,       1J,      2^,      1|1,      W 


VARIABLES  AND  LIMITS.  343 


Indeterminate  Forms. —Vanishing  Fractions. 

389.  When  one  or  more  variables  are  involved  in  both 
numerator  and  denominator  of  a  fraction,  it  may  happen  that 
for  certain  values  of  the  variables  both  numerator  and  denom- 
inator of  the  fraction  vanish.      The  fraction  then  assumes 

the  indeterminate  form  -  •     If  there  is  but  one  variable 

involved,  we  may  obtain  a  definite  value  as  follows : 

Let  x  be  the  variable,  and  a  the  value  of  x  "for  which  the 

fraction  assumes  the  form--     Give  to  a;  a  value  a  little 

greater  than  a,  as  a  +  h ;  the  fraction  will  now  have  a  defi- 
nite value.  The  limit  of  this  last  value,  as  h  is  indefinitely 
decreased,  is  called  the  limiting  value  of  the  fraction. 

The  fundamental  indeterminate  form  is  ->  and  all  other 

indeterminate  forms  may  be  reduced  to  this. 

rrn  on       a       b      a      0      0 

^  i  =  o"o  =  ox*=o' 

_       a      0  X  a      0 
0xco  =  ox-  =  — =  ~ 

a       b      0  X  a  —  0  X  S      0 

°°-00  =  o"o  = o =  0' 


1.    Find  the  limiting  value  of as  x  approaches  a. 

When  x  has  the  value  a,  the  fraction  assumes  the  form  -  •      » 
Put  x  =  a  +  h ;  the  fraction  becomes 

(a  +  h)  —  a  h 

Since  h  is  not  0,  we  can  divide  by  h  and  obtain  2  a  +  h. 
As  A  is  indefinitely  decreased,  this  approaches  2  a  as  a  limit. 


344  VARIABLES  AND  LIMITS. 

2.   Find  the  limiting  value  of      x         X2 — -  when  x 

ox   -|-  Z x   —  1 

becomes  infinite. 

Divide  each  term  of  the  numerator  and  denominator  by  x8.    Then, 

2x*-4x  +  5_       -x2      x« 
3x3  +  2x2  —  1  ~  2       1  ' 

O  T  1 

X        X3 

As  x  increases  indefinitely,  each  term  that  contains  x  of  the  last 

fraction  approaches  0  as  a  limit  (Theorem  4),  and  the  fraction  ap- 

2 

proaches  -  as  a  limit. 

o 

Exercise  130. 
Find  the  limiting  value  of : 

1.  ^  _   a — Q — ,    .   '  when  x  becomes  infinitesimal. 

7  x8  —  6  x  +  4 

0     fo2-5)(s2  +  7) 

2.  ■* 4  \;QK -  when  a;  becomes  infinite. 

X    -|~  oo 

(a;  4-  2)8 

3«       o,!   when  a;  becomes  infinitesimal. 
a;2  +  4 

4-   a,2_7a,  +  12  wnen  -  approaches  3. 

x2  —  9 
6*   s2  +  9  x  +  18  When  X  aPProaches  ~  3- 

6'  tt»  +  3s«  +  5s  +  3  When  X  aPProaches  "  *- 

7>   s»+*2*2-2a;-l  When  *  aPProaclies  L 


a4aj  +  Vaj— 1,  ..  .  „ 

8.  .  when  x  approaches  1. 

2x-  V«+l 


CHAPTER  XXIV. 
PROPERTIES  OF  SERIES. 


Convergent  and  Divergent  Series. 

390.  By  performing  the  indicated  division,  we  obtain 

from  - the  infinite  series  1  +  x  +  x2  +  xz  + This 

1  —  x 

series,  however,  is  not  equal  to  the  fraction  for  all  values  of  x. 

391.  If  x  is  numerically  less  than  1,  the  series  is  equal  to 
the  fraction.  In  this  case  we  can  obtain  an  approximate 
value  for  the  sum  of  the  series  by  taking  the  sum  of  a 
number  of  terms ;  the  greater  the  number  of  terms  taken, 
the  nearer  will  this  approximate  sum  approach  the  value  of 
the  fraction.  The  approximate  sum  will  never  be  exactly 
equal  to  the  fraction,  however  great  the  number  of  terms 
taken ;  but  by  taking  enough  terms  it  can  be  made  to  differ 
from  the  fraction  by  as  little  as  we  please. 

Thus,  if  x  =  -y  the  value  of  the  fraction  is  2,  and  the 

.      .  Ill 

series  is  1+-  +  -.  +  -+ 

The  sum  of  four  terms  of  this  series  is  l£ ;  the  sum  of 
five  terms,  ljj;  the  sum  of  six  terms,  1§£;  and  so  on. 
The  successive  approximate  sums  approach,  but  never 
reach,  the  finite  value  2. 


An  infinite  series  is  said  to  be  convergent  when  the 
sum  of  the  terms,  as  the  number  of  terms  is  indefinitely 
increased,  approaches  some  fixed  finite  value  ;  this  finite 
value  is  called  the  sum  of  the  series. 


346  PROPERTIES    OF  SERIES, 

393.  In    the    series    1  +  x  +  x2  +  xs  + suppose  x 

numerically  greater  than  1.  In  this  case  the  greater  the 
number  of  terms  taken,  the  greater  will  their  sum  be ;  by 
taking  enough  terms  we  can  make  their  sum  as  large  as 
we  please.  The  fraction,  on  the  other  hand,  has  a  definite 
value.  Hence,  when  x  is  numerically  greater  than  1,  the 
series  is  not  equal  to  the  fraction. 

Thus,  if  x  =  2,  the  value  of  the  fraction  is  —  1,  and  the 
series  is 

1  +  2  +  4  +  8  + 

The  greater  the  number  of  terms  taken,  the  larger  the  sum. 
Evidently,  the  fraction  and  the  series  are  not  equal. 

394.  In  the  same  series  suppose  x  =  1.     In  this  case  the 

1  1 

fraction  is =  -,  and  the  series  1+1+1+lH ■ 

The  more  terms  we  take,  the  greater  will  the  sum  of  the 
series  be,  and  the  sum  of  the  series  does  not  approach  a 
fixed  finite  value. 

If  x,  however,  is  not  exactly  1,  but  is  a  little  less  than  1, 

the  value  of  the  fraction will  be  very  great,  and  the 

JL        X 

fraction  will  be  equal  to  the  series. 

Suppose  x  =  —  1.     In  this  case  the  fraction  is  .       .  =  - , 

1  +  1       Z 

and  the  series  1  —  1  +  1  —  1  + If  we  take  an  even 

number  of  terms,  their  sum  is  0 ;  if  an  odd  number,  their 

sum  is  1.     Hence,  the  fraction  is  not  equal  to  the  series. 

395.  A  series  is  said  to  be  divergent  when  the  sum  of  the 
terms,  as  the  number  of  terms  is  indefinitely  increased, 
either  increases  without  end,  or  oscillates  in  value  without 
approaching  any  fixed  finite  value. 


PROPERTIES   OF  SERIES.  347 

No  reasoning  can  be  based  on  a  divergent  series ;  hence, 
in  using  an  infinite  series  it  is  necessary  to  make  such 
restrictions  as  will  cause  the  series  to  be  convergent.    Thus, 

we  can  use  the  infinite  series  1  +  x  +  x2  +  x8  + when, 

and  only  when,  x  lies  between  -f  1  and  —  1. 

396.  Theorem.  If  two  series,  arranged  by  powers  ofx, 
are  equal  for  all  values  of  x  that  make  both  series  con- 
vergent, the  corresponding  coefficients  are  equal  each  to  each. 


For,  if  A  +  Bx  +  Cx2  + =  Af  +  B'x  +  Cx2  +    •  ., 

by  transposition, 

A-Af  =  (B'-B)x  +  (C -  C)x2  +  ..... 

Now,  by  taking  x  sufficiently  small,  the  right  side  of  this 
equation  can  be  made  less  than  any  assigned  value  what- 
ever, and  therefore  less  than  A  —  A1,  if  A  —  Ar  has  any 
value  whatever.     Hence,  A  —  A'  cannot  have  any  value. 

Therefore,         A  -  A'  =  0,  or  A  =  A\ 

Hence,  Bx  +  Cx2  +  Dx*  + =  B'x  +  Cx2  +  D'x*  + , 

or         (B-BT)x  =  (C  -  C) x2  +  (D'  -D)x*  +  ..... 
Divide  by  x, 

B-B'  =  (C'-C)x  +  (£>'  -  D)x2  +  ...» 
By  the  same  proof  as  for  A  —  A\ 

B-B'  =  0,otB  =  BK 
In  like  manner, 

C=  C,  D  =  Df',  and  so  on. 
Hence,  the  equation 

A  +  Bx+  Cx2  + =  A'  +  B'x  +  C'x2  +  »..., 

if  true  for  all  finite  values  of  x,  is  an  identical  equation ; 
that  is,  the  coefficients  of  like  powers  of  x  are  equaL 


348  PROPERTIES   OF  SERIES. 

Indeterminate  Coefficients. 

397.   Expand  — - — — —  in  ascending  powers  of  x. 

JL  ~t~  35  "i    »C 

Assume        ■  ^+  ^x  ,  =  A  +  Bx  +  Cx*  +  Dx»  + ; 

1  +  X  +  x* 

then,  by  clearing  of  fractions, 

2  +  Sx  =  A  +  Bx  +  Cx*  +  Dx*  + 

+  Ax+  Bx*  +  Cxs  + 

+  Ax*  +  Bx*  + 

.:2  +  3x  =  A  +  (B  +  A)x  +  (C  +  B  +  A)x*  +  (D+C+B)x*  + 

By  §  396,      A  =2,  B  +  A  =  3,  C  +  B  +  A-  0,  D+C  +  £  =  0; 
whence,  2?  =  1,  C  =  —  3,  D  =  2 ;  and  so  on. 

2  +  3* 


'  *  1  +  X  +  *2 


=  2  +  x  -  3  x2  +  2  x3  + 


The  series  is  of  course  equal  to  the  fraction  for  only  such  values  of 
x  as  make  the  series  convergent. 

Note.  In  employing  the  method  of  Indeterminate  Coefficients, 
the  form  of  the  given  expression  must  determine  what  powers  of  the 
variable  x  must  be  assumed.  It  is  necessary  and  sufficient  that  the 
assumed  equation,  when  simplified,  shall  have  in  the  right  member 
all  the  powers  of  x  that  are  found  in  the  left  member. 

If  any  powers  of  x  occur  in  the  right  member  that  are  not  in  the 
left  member,  the  coefficients  of  these  powers  in  the  right  member  will 
vanish,  so  that  in  this  case  the  method  still  applies ;  but  if  any  powers 
of  x  occur  in  the  left  member  that  are  not  in  the  right  member,  then 
the  coefficients  of  these  powers  of  x  must  be  put  equal  to  0  in  equating 
the  coefficients  of  like  powers  of  x ;  and  this  leads  to  absurd  results. 
Thus,  if  it  were  assumed  that 

2  +  3x 


1+X+X2 


=  Ax  +  Bx*  +  Ox3  + 


there  would  be  in  the  simplified  equation  no  term  on  the  right  cor- 
responding to  2  on  the  left ;  so  that,  in  equating  the  coefficients  of 
like  powers  of  x,  2,  which  is  2x°,  would  have  to  be  put  equal  to  0x°; 
that  is,  2  =  0,  an  absurdity. 


PROPERTIES   OF  SERIES.  349 


10. 


11. 


Exercise  131. 

Expand  to  four  terms  : 

1 

1-x 

7. 

3  +  x 

*    1  +  2* 

™    1+x  +  x2 

1  —  x  —  X2 

1 

5-2x 
&'   1+x-x2 

8. 

1+x 

'   2-3x 

1+  X  +  X2 

1+x 

2—3a 

9. 

l-8z 

'   2  +  3a 

l-2x  +  3x2 

1  —  x-6x2 

Expand  to  five  terms : 

4 

12        5-20. 

14. 

3x-2 

'    2+x 

A~*    l+3a-z2 

x(x-iy 

2  -x 
'   3+x' 

aj(a?-2) 

15. 

x2-x  +  l 
(*-l)(x2  +  l) 

Partial  Fractions. 

398.  To  resolve  a  fraction  into  'partial  fractions  is  to 
express  it  as  the  sum  of  a  number  of  fractions  of  which 
the  respective  denominators  are  the  factors  of  the  denom- 
inator of  the  given  fraction.  This  is  the  reverse  of  the  proc- 
ess of  adding  fractions  that  have  different  denominators. 

Resolution  into  partial  fractions  may  be  easily  accom- 
plished by  the  use  of  indeterminate  coefficients  and  the 
theorem  of  §  396. 

In  decomposing  a  given  fraction  into  its  simplest  partial 
fractions,  it  is  important  to  determine  what  form  the  assumed 
fractions  must  have.  Since  the  given  fraction  is  the  sum  of 
the  required  partial  fractions,  each  assumed  denominator 
must  be  a  factor  of  the  given  denominator ;  moreover,  all 
the  factors  of  the  given  denominator  must  be  taken  as 
denominators  of  the  assumed  fractionso 


350  PROPERTIES    OF  SERIES. 

Since  the  required  partial  fractions  are  to  be  in  their 
simplest  form,  incapable  of  further  decomposition,  the  nu- 
merator of  each  required  fraction  must  be  assumed  with 
reference  to  this  condition.  Thus,  if  the  denominator  is 
xn  or  (x  ±  a)n,  the  assumed  fraction  must  be  of  the  form 

A  A  '  A  Ax  +  B       Ax  +  B 

—  or  - — ■ — — :  for,  if  it  had  the  form —  or — > 

xn       (x±a)ni        '  xn  (x±.a)n 

it  could  be  decomposed  into  two  fractions,  and  the  partial 

fractions  would  not  be  in  the  simplest  form  possible. 

When  all  the  monomial  factors,  and   all  the   binomial 

factors,  of  the  form  x  ±  a,  have  been  removed  from  the 

denominator  of   the   given  expression,  there  may  remain 

quadratic   factors   that   cannot   be   further  resolved;   and 

the  numerators  corresponding   to   these  quadratic  factors 

may  each  contain  the  first  power  of  x,  so  that  the  assumed 

fractions  must  have  either  the  form  -= —7   or   the 

ar  ±  ax  +  0 

.        Ax  +  B 
form  • 

x2  +  b 


3 

1.   Resolve  into  partial  fractions. 

X    ~\~  JL 

Since  x8  +  1  =  (x  +  1)  (x3  —  x  +  1),  the  denominators  will  be  x  +  1 
and  x2  —  x  +  1. 

3  A      1      Bx  +  C 

Assume  -ttt  =  — r-:  + 


x8  +  l      x  +  lxs-x  +  l' 
then,  3  =  A(x*  -  x  +  1)  +  (Bx  +  C)  (x  +  1) 

=  (A  +  B)x*  +(B+C-A)x  +  (A  +  C); 
whence,  3  =  A  +  C,  B+  C-A  =  0,  A  +  B  =  0t 

and  -4«=1,  B=  — 1,  (7  =  2. 


Therefore, 


3  _1 x-2 

x8  +  l      x  +  1      x3-x  +  l 


PROPERTIES   OF  SERIES.  351 


2.    Kesolve 2 into  partial  fractions. 


4 xs  -x2  -Sx  -2  . 

a2  (x  4- 1) 

The  denominators  may  be  x,  x2,  x  4-  1,  (x  +  l)2 


4x8-x2-3x-2      ^L  ,  B   ,       C 
Assume  tz — r~ 777 =  — K  +  M  ■   ,  + 


x2  (x  +  l)2  x       x2      x  +  1      (x  +  l)2 

.-.  4x3  - x2  -  3x  -  2  =Ax{z  +  l)2  +B{x  +  l)2  +  Cx2(x  +  1)+Dxa 
=  (^  +  C)x8  +  (2^  +  B  +  C  +  D)x2  +  (^  +  2B)x  +  JB; 
whence,  A  +  C  =  4, 

A  +  2£  =  -3, 
B  =  -2; 
and  .-.  B  =  -  2,   ^i  =  1,    C  =  3,   D  =  -  4. 

Therefore    4x«-x2-3x-2=:l_2      _3 4       ■ 

lnerelore,  JC2(x+1)2  x      x2  +  x  +  1      (x  +  1)* 

Exercise  132. 
Kesolve  into  partial  fractions  : 

1. 


7z  +  l 


4. 


5. 


(x  +  4)  (x  -  5) 

7a-l 
(l-2z)(l-3a:) 

5x-l 

(2x-l)(x-5)' 

x-2 
(x-5)(x  +  2)' 

3 


x2  -  x  -  3 
6. 


7. 

3ce2-4 

x2  (x  +  5) 

8. 

7cc2-a: 

(x-l)2(z-f-2) 

9. 

2z2-7a  +  l 

*8-l 

7  a;-  1 

1U. 

(6ic-hl)(o;-l) 

11. 

a2 -3 

(a2  +  1)  (x  +  2) 

1 0 

x2-x  +  l 

x  (x2  -  4)  (x2  +  !)(»-  I)5 


CHAPTER   XXV. 
BINOMIAL  THEOREM. 

399.  Binomial  Theorem,  Positive  Integral  Exponent.  By  suc- 
cessive multiplication  we  obtain  the  following  identities : 

(a  +  by  =  a*  +  2ab  +  b*; 

(a  +  b)8  =  a*  +  3  a%  +3ab2  +  b8-, 

(a  +  by  =  a*  +  4a*b  +  6a2b2  +  4a*8  +  b\ 

The  expressions  on  the  right  may  be  written  in  a  form 
better  adapted  to  show  the  law  of  their  formation : 

(a  +  by  =  a*  +  2ab  +  |4&V 

/    _i_*\«        4.j    8a.  4'3    2?2  ,  4-3-2    ,„      4-3-2-lr, 

Note.  The  dot  between  the  Arabic  figures  means  the  same  as  the 
sign  X. 

400.  Let  n  represent  the  exponent  of  (a  -f  b)  in  any  one 
of  these  identities;  then,  in  the  expressions  on  the  right, 
we  observe  that  the  following  laws  hold  true : 

1.  The  number  of  terms  is  n  +  1. 

2.  The  first  term  is  an,  and  the  exponent  of  a  is  one  less 
in  each  succeeding  term. 

3.  The  first  power  of  b  occurs  in  the  second  term,  the 
second  power  in  the  third  term,  and  the  exponent  of  b  is 
one  greater  in  each  succeeding  term. 

4.  The  sum  of  the  exponents  of  a  and  b  in  any  term  is  ft. 


BINOMIAL    THEOREM.  353 

5.  The  coefficients  of  the  terms  taken  in  order  are  1 ;  n ; 
»» *-■!):  n(n-l)(n-2)         d 

1-2      '  1-2-3  '   anasoon- 

401.  Consider  the  coefficient  of  any  term ;  the  number 
of  factors  in  the  numerator  is  the  same  as  the  number  of 
factors  in  the  denominator,  and  the  number  of  factors  in 
each  is  the  same  as  the  exponent  of  b  in  that  term ;  this 
exponent  is  one  less  than  the  number  of  the  term. 

402.  Proof  of  the  Theorem.  That  the  laws  of  §  400  hold 
true  when  the  exponent  is  any  positive  integer  is  shown 
as  follows : 

We  know  that   the   laws   hold   for  the  fourth   power; 
suppose,  for  the  moment,  that  they  hold  for  the  &th  power. 
We  shall  then  have 

(a  +  b)k  =  a*  +  ka*~lb  +  ^"^  a*-*b* 

+  *(*-l)(*-2)at_,&,+ (1) 

Multiply  both  members  of  (1)  by  a  +  b ;  the  result  is 
(a  +  b)*  +  *  =  ak  +  l  +  (k  +  l)  a*b  4-  QL+3^#~ip 

+  (*+i)*(*-iW+ (2) 

In  (1)  put  h  -b  1  f or  h ;  this  gives 
(a  +  b)*'+l  *=a*+*  +  (k  + 1)  a*b  +  (fr+'iK* + 1  -  *)  ak-ip 

+  (*  +  !)(*  +  !  -1)^  +  1-2)  ak_2b8  + 

1  *  Z  '  o 

=  a**1  +  (k  +  1)  aH  +  (k  +  W  ak-ih2 

+(.+i)M.-i)^,+ (3) 

Equation  (3)  is  seen  to  be  the  same  as  equation  (2). 


354  BINOMIAL    THEOREM. 

Hence,  (1)  holds  when  we  put  k  -f- 1  for  k ;  that  is,  if  the 
laws  of  §  400  hold  for  the  kth.  power,  they  must  hold  for 
the  (k  +  l)th  power. 

But  the  laws  hold  for  the  fourth  power  ;  therefore,  they 
must  hold  for  the  fifth  power. 

Holding  for  the  fifth  power,  they  must  hold  for  the  sixth 
power ;  and  so  on  for  any  positive  integral  power. 

Therefore,  they  must  hold  for  the  wth  power,  if  n  is  a 
positive  integer ;  and  we  have 


(a  +  b)n  =  an  +  na*~lb  +  %&    ^  a* 

1  •  J!i 


+  re(ra-l)(r2W+ .  (A) 

Note.  The  above  proof  is  an  example  of  a  proof  by  mathematical 
induction.     See  §  134. 

403.  This  formula  is  known  as  the  binomial  theorem. 

The  expression  on  the  right  is  known  as  the  expansion  of 
(a  -f-  b)n ;  this  expansion  is  a  finite  series  when  n  is  a  positive 
integer.     That  the  series  is  finite  may  be  seen  as  follows : 

In  writing  out  the  successive  coefficients  we  shall  finally 
arrive  at  a  coefficient  that  contains  the  factor  n  —  n ;  and, 
therefore,  this  term  will  vanish.  The  coefficients  of  all  the 
succeeding  terms  likewise  contain  the  factor  n  —  n,  and, 
therefore,  all  these  terms  will  vanish. 

404.  If  a  and  b  are  interchanged,  the  identity  (A)  may 
be  written 

(a  +  b)n  =  (b  +  a)n  =  bn  +  nb^a  +  W^~1V~V 

.  n  (n  —  1)  (n  —  2)  .     .  R  , 
+     K     1-2-3 lb"-'a'  + 


BINOMIAL    THEOREM.  355 

This  last  expansion  is  the  expansion  of  (A)  written  in 
reverse  order.  Comparing  the  two  expansions,  we  see 
that  the  coefficient  of  the  last  term  is  the  same  as  the 
coefficient  of  the  first  term;  the  coefficient  of  the  last  term 
but  one  is  the  same  as  the  coefficient  of  the  first  term  but 
one;  and  so  on. 

In  general,  the  coefficient  of  the  rth  term  from  the  end 
is  the  same  as  the  coefficient  of  the  rth  term  from  the 
beginning.  In  writing  an  expansion  by  the  binomial 
theorem,  after  arriving  at  the  middle  term,  we  can  shorten 
the  work  by  observing  that  the  lemaining  coefficients  are 
those  already  found,  taken  in  reverse  order. 

405.  If  b  is  negative,  the  terms  that  involve  even  powers 
of  b  will  be  positive,  and  the  terms  that  involve  odd  powers 
of  b  will  be  negative.     Hence, 

(a  -  b)n  =  an-  na*-*b  +  n(\~V)  a»-%* 

1  *  2i 

n(n  —  l)(n  —  2)        ...  rtr. 
— 1-2-3 + W 

Also,  putting  1  for  a  and  x  for  b  in  (A)  and  (B). 

(1  +  x)n  =  1  +  nx  4-  n^l~V)xi 

+^~i1}2(.r2)*8+ <c) 


(1  -  x)«  =  1  -  nx  +  *fe    -j&a* 


1-2 


_n(n-l)Jn-2)xi+ ^ 


1-2-3 


356  BINOMIAL    THEOREM. 

1.   Expand  (1  +  2x)\ 

In  (C)  put  2  x  f or  x  and  5  for  n.     The  result  is 

(1  +  2x)«  =  1  +  5(2x)  +  f^f  4*2  +  rfr|8x8 

+  l-2-3-416iC  +1-2-3-4-6323* 
=  1  +  10s  +  40*2  +  80*8  +  80x*  +  32  gfi. 


2.   Expand  to  three  terms 


/l  _  2£>y 


. ... 


1  2x2 

Put  a  for  - ,  and  6  for  — ;  then,  by  (B), 
x  <-> 

(«  -  6)6  =  a6  _  6  a5&  +  15  a4&2  _ 

Replace  a  and  6  by  their  values, 
x6     x»  +  3 


406.  Any  Required  Term.  From  (A)  it  is  evident  (§  400) 
that  the  (r  +  l)th  term  in  the  expansion  of  (a  +  b)n  is 

7t  (n  —  1)  (n  —  2) to  r  factors    n_r,r 

1x2x3 r  a 

Note.     In  finding  the  coefficient  of  the  (r  +  l)th  term,  write  the 

series  of  factors  1  X  2  X  3 r  for  the  denominator  of  the  coefficient, 

then  write  over  this  series  the  factors  n  (n  —  1)  (n  —  2),  etc.,  writing 
just  as  many  factors  in  the  numerator  as  there  are  in  the  denominator. 

The  (r  -f  l)th  term  in  the  expansion  of  (a  —  b)n  is  the 
same  as  the  above  if  r  is  even,  and  the  negative  of  the 
above  if  r  is  odd. 


BINOMIAL    THEOREM.  357 


Find  the  eighth  term 


°'M)» 


x2 
Here  a  =  4,  b  =  — »  n  =  10,  r  =  7. 

3 

™    *  1    ^-    10-9-8-7-6-5-4/J%lt/      x2\* 

The  term  required  is  —- y— ^  (4)8 ( -  -)  , 

which  reduces  to  —  60  x14. 

407.   A  trinomial  may  be  expanded  by  the  binomial 
theorem  as  follows : 
Expand  (1  +  2  x  -  a2)8. 

Put  2  x  —  x2  =  z. 

Then,  (1  +  zf  =  1  +  3z  +  3*2  +  2«. 

Replace  z  by  2  a;  —  x2. 

.%  (1  +  2x-x2)3  =  1  +  3(2x-x2)  +  3(2x-x2)2  +  (2X-X2)8 
=  1  +  6x  +  9x2  —  4x3  —  9x4  +  6x6  —  x°. 

Exercise  133. 
Expand : 

1.  (a  +  b)\  6.    (a»  +  by.  II.    (m-J  +  w2)4. 

2.  (x-2)5.  ?>    (m«  +  w«)8#         12.    (ar-»  +  **)•. 

3.  (Sx  -2yY.  , 

8.    (a -by.  13.    (2x2  +  ^6. 

/  £C         "2/  1 

4'  \y"*y         9-  (**  +  *1)5-     l4-  (»*-cV. 

5.    (4  +  3y)4.  10.    (a-^  +  fl-2)8.      15.    (2  a2  -  £V^)5. 


18.    (2x2y-1-y^/y)4.        21.    (2^-2-^'t 


358  BINOMIAL    THEOREM. 

24.  Find  the  fourth  term  of  (2  a;  -  Sy)\ 

25.  Find  the  ninety-seventh  term  of  (2  a  —  b)100. 

Note.  As  the  expansion  has  101  terms,  the  ninety-seventh  term 
from  the  beginning  is  the  fifth  term  from  the  end. 

26.  Find  the  eighth  term  of  (3  x  —  y)l\ 

27.  Find  the  tenth  term  of  (2  a2  -  -J-a)20. 

28.  Find  the  fifth  term  of  (a  -  2  V&)25. 

29.  Find  the  eleventh  term  of  (2  —  a)16. 

30.  Find  the  fifteenth  term  of  (x  +  y)20- 

31.  Find  the  fourth  term  of  (3  -  2  ic2)9. 

32.  Find  the  twelfth  term  of  (a?  -  a  Vic)17. 

33.  Find  the  seventh  term  of  (y2  -  l)88. 

34.  Find  the  fifth  term  of  Qa-bVb)21. 

35.  Find  the  fourth  term  of  (VS  -  V^2)20. 

36.  Find  the  third  term  of  ( Va  —  V^~6)7. 

37.  Find  the  sixth  term  of  (Va3  -  V^)9. 

38.  Find  the  eighth  term  of  (Vfa  +  Vfa-)20. 

39.  Find  the  ninth  term  of  (x^T^T  +  y  V^I)16. 

a*b  -         -  >    • 


of  (t 


vw 

41.   Find  the  seventh  term  of  (x  •+•  aj_1)2n. 


BINOMIAL    THEOREM.  359 

408.  Binomial  Theorem,  Any  Exponent.  We  have  seen 
(§  402)  that  when  n  is  a  positive  integer  we  have  the 
identity 

(l+xy  =  l+nx  +  ^^x>  +  n{n-1l%-2h>+ 

We  proceed  to  the  case  of  fractional  and  negative  expo- 
nents. 

I.    Suppose  n  is  a  positive  fraction,  —  >  in  which  p  and 

q  are  positive  integers.     We  may  assume  that 

(i  +  xy  =  (A  +  Bx+  cx2  +  Dx*  + y,         (1) 

provided  x  is  so  taken  that  the  series 

A  +  Bx  -f  Cx*  +  Dx%  + 

is  convergent,  §  392. 

That  this  assumption  is  allowable  may  be  seen  as  follows : 
Expand  both  members  of  (1).     We  obtain 

1+^  +  £^,2  +  E(£z^|^JC3  + 9 

+ 


and  A«  +  qA«~lBx+  IfM     ^^Ai-'P+qA*-1^]  x 

In  the  first  k  coefficients  of  the  second  series  there  enter 

only  the  first  k  of  the  coefficients  A,  B,  C,  D, If,  then, 

we  equate  the  coefficients  of  corresponding  terms  in  the 
two  series  (§  396)  as  far  as  the  &th  term,  we  shall  have  just 
k  equations  to  find  k  unknown  numbers  A,  B,  C,  D9  •••% 
Hence,  the  assumption  made  in  (1)  is  allowable. 

Equating  the  two  first  terms  and  the  two  second  terms, 
we  obtain 

A*  =  1,        .\  A  =  1 ; 
qAq"1B=p1  or  qB=pf         .*.  J5  =  ^- 


360  BINOMIAL    THEOREM. 

Extracting  the  ^th  root  of  both  members  of  (1),  we  have 

(1  +  x)\  =  l  +^x  +  Cx2  +  Dx*  + ,  (2) 

where  x  is  to  be  so  taken  that  the  series  on  the  right  is 
convergent. 

II.    Suppose  n  is  a  negative   number,  integral  or  frac- 
tional.    Let  n  =  —  ra,  so  that  m  is  positive ;  then, 

1 


(1  +  x)n  =  (1+  x)~m 


(i  +  xy 


From  (2),  whether  m  is  integral  or  fractional,  we  may 
assume 

1        = 1 

(1+  x)m  ~  1  +  mx  +  cx2  +  efo8  + * 

By  actual  division  this  gives  an  equation  in  the  form 

(1  +  x)~m  =  l-mx  +  Cx2  +  Dx8  + (3) 

409.  It  appears  from  (2)  and  (3)  (§  408)  that  whether  n  is 
integral  or  fractional,  positive  or  negative,  we  may  assume 

(1  +  x)n  =  1  +  nx  4-  Cx2  +  Dx8  4- , 

provided  the  series  on  the  right  is  convergent. 

Square  both  members, 

(l  +  2x  +  x*)n  =  l  +  2nx  +  2Cx2  +  2Dx*  +  ••••• 

+  n2x2   +2nCx*  + (1) 

Also,  since 

(1  +  y)»  =  1  +  ny  4-  Cy2  4-  Dys  + , 

we  have,  putting  2  x  4-  x2  for  y> 

(1  4-  2  x  4-  xy  =  1  4-  n  (2  x  +  x2)  4-  C  (2  x  -\-  x2)2 

+  D(2x  \-x*)* 

=  l+2nx  +  nx2    +4Ccc84- 

+  4:Cx2+$Dx*  + (2) 


BINOMIAL    THEOREM.  361 

Equate  corresponding  coefficients  in  (1)  and  (2), 

n  +  ±C  =  2C  +  n2, 
±C  +  8D  =  2D  +  2nC. 

,'.  2  <7  =  »*-*,  and       C  =  W^1); 

3i>=(,-2)fiandi>^M^-^-^. 

and  so  on. 

Hence,  whether  n  is  integral  or  fractional,  positive  or 
negative,  we  have 

(1  +  *)-  =  1  +  nx  +  S^V  +  ^-l)fc-2)a,  + , 

provided,  always,  x  is  so  taken  that  the  series  on  the  right 
is  convergent. 

The  series  obtained  will  be  an  infinite  series  unless  n  is  a 
positive  integer,  §  403. 

410.   If  x  is  negative, 

/-.         n„      i  ,  n(n  —  l)  2      n(n  —  l)(n  —  2)   8 

(1  -  x)n  =  1  -  nx  +     v  V v     1-2-3 ~*~ 

Also,  if  x  <  a, 

.=anL1  +  "«+   1^2  s  + J 


if  x  >  a, 


=  a»  +  fl»r^  +  SfergSi^.^  + ; 

r      |  .   • .  -     •     \ 


(a  +  x)'  =(x  +  a)«  =  x»(l+^Y 

|_  #  12     x2  J 

*      -  s» ■+  nax*-1  +  "^"^aV-2  +  . 


362  BINOMIAL    THEOREM. 


1.   Expand  (1  +  x)*. 


The  above  equation  is  true  only  for  those  values  of  x  that  make 
the  series  convergent. 

2.  Expand  — • 

T==  =  (!-*>-* 

VI  —  X 

=  i-(-^+=tili^-~^1~f3~f»8+ 

if  x  is  so  taken  that  the  series  is  convergent. 

A  root  may  often  be  extracted  by  means  of  an  expansion* 

3.  Extract  the  cube  root  of  344  to  six  decimal  places. 

844  =  843(1  +  8i3)  =  78(1  +  3i3)- 

'[_        3\343>M       1-2      1,343/  +       J 

=  7  (1  +  0.000971817  -  0.000000944  + ) 

'       =  7.006796. 

4.  Find  the  eighth  term  of  f  x -=  )     . 

\         4Va?/ 

Here 


a 

=  x, 

4Vx 

8. R  .  _ 

?            7 

3 

— ,  n  = 

4x 

1 

T 

r=7. 

> 

1-2-3 
1-3-5-7 

4-5-6-7 
•9-1113 

37 

X 

1-3-5-7-9-11-13-37 
2-4-6-8-10-12-14-47-xu 


BINOMIAL    THEOREM.  363 

Exercise  134. 
Expand  to  four  terms : 

1.  (1+*)*  6.   (1+a)*  11.   (2a  +  3y)* 

2.  (1+tf)-2.  7.    (l+x)~\  12.    (2a*  +  3y)~* 

3.  (1 +*)"*.  8.    (1+z)-8.  13. 

Va2  —  cc* 

4.  (1-s)*  9.    (l  +  5aj)-5.  j 

14-      - 

5.  (1  -  x)~*'  10.    (1  +  5 a)1.  V(a  -  xf 


15.  Find  the  fourth  term  of 

16.  Find  the  fifth  term  of 


\        2>/xJ 


^/(a-2xy 

17.  Find  the  third  term  of  (4  -  7  x)\ 

18.  Find  the  sixth  term  of  (a2  —  2  axy. 

19.  Find  the  fifth  term  of  (1  -  2  xfK 

20.  Find  the  fifth  term  of  (1  -  x)~\ 

21.  Find  the  seventh  term  of  (1  —  x)  . 

! 

22.  Find  the  third  term  of  (1  -f  x)   2». 

23.  Find  the  fourth  term  of  (1  4-  x)~K 

24.  Find  the  sixth  term  of  (2 V  • 

25.  Find  the  fifth  term  of  (2  x  -  3  y)~*, 

26.  Find  the  fourth  term  of  (1  —  5x)~K 


364  MISCELLANEOUS  PROBLEMS. 

Exercise  135.  —  General  Be  view. 

1.  Add  (a  -  b)  x2  +  (b  -  c)  y2  +  (c  -  a)  z* ;      (b  -  c)  x2 

+  (e  -  a)  y2  +  (a  -  b)  z2;     (c  -  a)  x2  +  (a  -  b)  y2 
+  (b-c)z\ 

2.  Add(a  +  b)x  +  (b  +  c)y-  (c  +  a)z;  (b  +  c)z+  (c  +  a)x 

—  (a  +  b)y\  (a  +  c)y+  (a  +  b)z—  (b  +  c)x. 

3.  From  4a;8  —  §x2  +  %x  —  7  take  the  sum  of 

'Sx8  +  7  -  8a;2  +  7x  and  -  9  a;8  -  8x2  +  4a;  +  4. 

4.  Find  the  product  of  ap  —  Sal1'-1  +  ±ap~2  —  6ap~8  +  5ap~4 

and  2  a*  —  a2  +  a. 

5.  Divide  1  -  6x5  +  5a;6  by  1  -  2x  +  x2. 

6.  Divide  4  /**+1  -  30  fc*  +  19  A*-1  +  5  A*-2  +  9  A*~4 

by  h*-*  -  7  h*-*  +  2  7^-5  -  3  hx~\ 

7.  Simplify  3Ja  -  2  (J  -  c)  J  -  [45  +  \2  b  -  (c  -  a)}]. 

8.  Find  the  factors  of  10  x2  +  79  a;  -  8. 

9.  Find  the  H.C.F.  of  2a;8  +  a;2  +  4a;  -  7 

and  a;8- 2a;2 +  1. 

10.  Keduce  to  lowest  terms  ■»,,,>. — si  * 

5  a4  +  9  a8  —  64 

11.  If  a  =  4,  b  =  £,  c  =  —  2,  find  the  numerical  value  of 

C  £2 


12. 


T,  a +  5 ,        ^x/^- aY      a;  — 2a  +  &       A 

If  x  =  — ^r—  >  show  that  I  r ; ^77  =  0. 

2  \x  —  by        x-\-  a  —  2  b 


13.  If  a;  =  2  y  +  3  z,  show  that 

x*-8y*-27z*-  lSxyz  =  0. 

14.  Eesolve  into  factors  6  a;2  -f  5  a;  —  4. 

15.  Eesolve  into  factors  a;4  +  2  x%  -  13  x2  -  38  a;  -  24. 

16.  Eesolve  into  factors  12  a;8  4-  20  x2  —  x  —  6. 


MISCELLANEOUS  PROBLEMS.  365 

17.  A  boy  bought  a  number  of  apples  at  the  rate  of  5  for 
2  cents.  He  sold  half  of  them  at  the  rate  of  2  for  a  cent 
and  the  rest  at  the  rate  of  3  for  a  cent,  and  cleared  a  cent 
by  the  transaction.     How  many  did  he  buy  ? 

Find  the  H.C.F.  and  the  L.C.M.  of: 

18.  3x6-5x*  +  2,  and2a;6-5a;2  +  3.. 

19.  3x»  +  10x2  +  7x-2,  and3z8-f  13z2  +  17a;  +  6. 

20.  Ax4-  9x2  -f  6x  —  1,  and  6x*-  7  x2  -f  1. 

21.  x6  +  11  cc  -  12,  and  «6  +  11  xz  +  54. 

22.  2z8  +  5a2y-5iC2/2  +  2/8,  and  2 a;8 - 7 x2y  +  5 a^2 - y\ 
Simplify : 

x  +  1  3aj  +  2     ,      2a;-l 


23. 


x(x  —  2)      x  (x  +  1)  ^  a;2 


_  ■    1+aj   .   1-z       1  +  x*      1  -  x* 
24. f- 


1-x  '   l-\-x      1-a:2      1+a;2 

2  s-1 2(x  +  2)  x  +  5 

'    (x  +  2)(x  +  5)       (x  +  fytx-l)^  (x-l)(x  +  2) 

27. 


ax  —  a2      ax  -\-2a2      x2  -\-  ax  —  2  a2 

x  -  1  a;  —  3 


(a +  3) (3-1)       (a> +  3) (2 -a)       (2-x)(l-x) 

I  +  i  +  l  t.i-ia-*) 

a       o       c  „  2 


28-T-T7T-T- 

«»      I       7.0  O  "T~ 


« 


2     '     7,2  ..2 


£2      c2  '  ab 

1  X  i  +  — 


A^     1  —  a      1  +  a  .  a      a£8 

29.    - 31. 


*-i+l 


1  —  a      1  4-  a  b 


366 


MISCELLANEOUS   PROBLEMS. 


32. 


33. 


1  + 


1  + 


1  + 


4  —  x 


l  +  a  + 


2  a? 
1+a 


34.  Two  passengers  have  together  400  pounds  of  bag- 
gage. One  pays  .$1.20,  the  other  $1.80,  for  excess  above  the 
weight  allowed.  If  all  the  baggage  had  belonged  to  one 
person,  he  would  have  had  to  pay  $4.50.  How  much  bag- 
gage is  allowed  free  ? 


Solve : 


37. 


oc     6 a; +  13      9z  +  15  .   0      2x  +  15 

35.     3~ — j;^  +  o  — ' 


15 


5  x  —  25 


2x  +  a    ,    Sx 
36.   7tz r  + 


S(x-a)      2(x  +  a) 


V 


=  41 


1+^  =  2* 

y 


6 


=  2J. 


OA    2x  t  3y      kz       . 

39. \--f- =  1 

a         o  c 

a         b  c 

a         b  c 


40. 


a 

b 

e 

-  + 

-  + 

= 

3 

iC 

y 

z 

o   . 

b 

c 

-  + 

ss 

1 

X 

y 

z 

2a 

b 

c 

— 

—  _ 



= 

0 

X 

y 

z 

Find  the  arithmetical  value  of : 

41.  36*;  27*;  16*;  32*;  4f;  81;  27*;  64*. 

42.  32*;   64*;   81*;    (3#j    (5TV)*;    (l*)1. 


MISCELLANEOUS  PROBLEMS.  367 

43.  (0.25)1;   (0.027)*;  49"  *;  32~f;   81~f. 

44.  36"*;  27~*;  (A)"*»   C0-116)"1;   (0.0016)-*. 

Arrange  in  ascending  order  of  magnitude : 
45.    6V7;  9V3;  5VlO.  46.   4V6;   3^3;  5^2. 

Simplify : 

/r  /=  /r  2VlO      7V48        4Vl5 

47.    *V3  X  4 VO -t-4. VS.       48.    7=  X =  -fc 7==" 

3  *  7  3V27      5Vl4      15V21 

49.    =  •     50.   =•     51.    = 52. 


2  +  V3  '    2-V3  '    2V5  +  3  !   2V5-3 

1  1 

53.    {C*W)fX  (a56)-2p«-2.  54.   (a;18oXa;-12)3«-2.  . 

55.  3  (a*  +  ft*) 2  -  4  (a*  +  ft*)  (a*  -  ft*)  +  (a*  -  2  ft4) 2. 

56.  {(a")*-m}S7i.  58.    [\(a-m)-n\*>y+[\(am)n\-P~\-* 

57     fxP  +  9\p  .   /   £j  V~a  £c2iJ(?-i)_^2</(p-i) 

\  a*  /    +  \a«"V      *  59'     iCp(3-1>+^^-1>  ' 

60.  5^320  -  2^-  1715  +  3^^135. 

61.  2V18-3V8  +  2V50;    -^81  +  v"24  -  J/192. 

62.  |Vj  + V80-iV20;    8V^  +  10 Vf  -  2Vj. 

Extract  the  square  root  of : 

63.  9ar4-18a-8y*  +  15a;-V-6ar-1^  +  2A 

Extract  the  cube  root  of : 

64.  8a*  4-  12a2  -  30a  -  35  +  45a-1  +  27a"2  -  27a"8. 


368  MISCELLANEOUS  PROBLEMS. 

Resolve  into  prime  factors  with  fractional  exponents: 
65.    -^12,  ^72,  -n/^.V'oT;    and  find  their  product. 
Solve: 

66. 


67. 


68. 


69. 


70. 


5 

x  -2 

4         3 

£C         03+6 

71. 

6  c2 

ace2 —  =  c#  —  6cc2. 

a  +  b 

jc  +  3 

2z-l 

72. 

3xy-5if  =    ll 

2«-7 

a-3   "a 

5xy  +  3x2  =  22) 

x+4 
x  —  4 

?  ~  2  -  fiJL 
•cc-3-6*' 

73. 

74. 

x2  +  10xy  =  ll] 
5xy-3y2  =    2} 

x  —  a 

Va  +  y  =  Vy  +  2  1 

cc  —  6 

x  —  a          ab 

x-y=7 ,               J 

ace  4-  b 

ex  -\-  d 

75. 

a2  +  a;i/  +  2/2  =  52l 

a  -\-bx 

c  +  dx 

icy  —  tc2  =  8 

Form  the  equation  of  which  the  roots  are  : 

76.  a  -  5,  a  +  £.  79.    1  +  VS,  1  -  V5. 

77.  a -26,  a +  36.  80.    -  1  +  V3,  -  1  -  VS. 

78.  a +  26,  2a +  6.  81.    1 -f  V^,  1  -  V^3. 

82.  A  vessel  that  has  two  pipes  can  be  filled  in  2  hours 
less  time  by  one  than  by  the  other,  and  by  both  together 
in  1  hour  52  minutes  30  seconds.  How  long  will  it  take 
each  pipe  alone  to  fill  the  vessel  ? 

83.  A  number  is  expressed  by  two  digits,  the  second  of 
which  is  the  square  of  the  other,  and  when  54  is  added  its 
digits  are  interchanged.     Find  the  number. 

84.  Divide  35  into  two  parts  such  that  the  sum  of  the 
two  fractions  formed  by  dividing  each  part  by  the  other 
may  be  2^. 


MISCELLANEOUS  PROBLEMS.  369 

85.  A  number  consists  of  three  digits  in  arithmetical 
progression ;  and  this  number  divided*  by  the  sum  of  its 
digits  is  equal  to  26 ;  but  if  198  is  added  to  the  number, 
the  digits  in  the  units'  and  hundreds'  places  will  be  inter- 
changed.    Find  the  number. 

i  86.  The  sum  of  the  squares  of  the  extremes  of  four  num- 
bers in  arithmetical  progression  is  200,  and  the  sum  of  the 
squares  of  the  means  is  136.     What  are  the  numbers  ? 

87.  Show  that  if  any  even  number  of  terms  of  the  series 

1,  3,  5  is  taken,  the  sum  of  the  first  half  is  to  the  sum 

of  the  second  half  in  the  ratio  1  :  3. 

88.  If  a  horse  took  1  second  more  for  each  rod,  he  would 
travel  1|  miles  less  per  hour.     Find  his  rate  of  traveling. 

Solve : 

89.  x8  +  y3  =  l$V2]  90.    2x2  +  Sxy  = 


z  +  2/  =  3V2 


1  90.    2x2  +  3xy  =  8^ 

J  y*-2xy  =  20  J 


Expand : 

91.  (x-x-y.     93.   (2ai-a~iy.     95.   QVx' -  i^x)\ 

92.  (^--£z)4.  94.  (2x-2  +  x*y.       96.   (jV^F3  +  ^ar1)8. 

97.  If  a,  b,  e,  d  are  in  continued  proportion,  prove  that 
b  +  c  is  a  mean  proportional  between  a  +  b  and  c  -+-  d. 

98.  If  a  +  b  :  b  +  c  =  c  +  d :  d  +  a, 

prove  that  a  =  c,  OYa  +  b-\-c-\-d  =  0. 

99.  The  number  of  eggs  which,  can  be  bought  for  1 
dollar  is  equal  to  twice  the  number  of  cents  which  32  eggs 
cost.     How  many  eggs  can  be  bought  for  1  dollar  ? 

100.   Find  two  fractions  whose  sum  is  ^¥,   and  whose 
difference  is  equal  to  their  product. 


370  MISCELLANEOUS  PROBLEMS. 

101.  The  velocity  of  a  falling  body  varies  as  the  time 
during  which  it  has  fallen  from  rest,  and  the  velocity  at 
the  end  of  2  seconds  is  64  ft.  Find  the  velocity  at  the 
end  of  6  seconds. 

102.  The  distance  through  which  a  body  falls  from  rest 
varies  as  the  square  of  the  time  it  falls ;  and  a  body  falls « 
144  ft.  in  3  seconds.     How  far  does  it  fall  in  4  seconds  ? 

103.  The  volume  of  a  gas  varies  directly  as  the  absolute 
temperature  and  inversely  as  the  pressure.  If  the  volume 
of  a  gas  is  1  cubic  foot,  when  the  pressure  is  15  and  the 
temperature  280,  what  will  be  the  volume  when  the  pres- 
sure is  35  and  the  temperature  320  ? 

104.  The  difference  between  the  first  and  second  of  four 
numbers  in  geometrical  progression  is  96  \  the  difference 
between  the  third  and  fourth  is  6.     Find  the  numbers. 

105.  If  «2,  b2,  c2  are  in  arithmetical  progression,  prove 
that  b  +  c,  c+a,  a  +  b  are  in  harmonical  progression. 

When  x  =  oo,  and  when  x  =  0,  find  the  limit  of : 

(2x  -3)  (3  -5x)  x*-x  +  l      , 

1Ub*        7sa-6aj  +  4  107,    (z2  +  l)(*-l)a 

Kesolve  into  factors  and  find  all  the  values  of  x : 

108.  a;4-5sc2  +  4  =  0.  114.  3^-5-^  +  2  =  0. 

109.  x6  -9  xs  +  8  =  0.  115.  6^-3^-45  =  0. 

110.  9o;4-13a:2  +  4  =  0.  116.  21  Va-2-  5V«  -  74=0. 

111.  4z4  -17a;2  +  4  =  0.  117.  3 Vx5  +  4^  -  20  =  0. 

112.  2a4-5a;2  +  2  =  0.  118.  2x«  -  19a8  +  24  =  0. 

113.  2^-3x^  +  1=0.  119.  a4- 1=0. 


§  MISCELLANEOUS  PROBLEMS.  371 

120.  «6-l=0.  124.  4af  *-3af^-27=0. 

121.  xs  +  8  as*  -  9  =  0.  125.  z2n  +  3  xn  -  4  =  0. 

122.  16^-17^  +  1  =  0.  126.  3x*  -2ax^  -a2  =  0. 

123.  x~*  +  5of*  -14  =  0.  127.  V2a- -  V^2»  -  2  =  0. 
Solve : 


128.    Va;  +  4  +  V3£c  +  l=V9z  +  4. 


129.    V5  x  +  1  +  2  V4z  -  3  =  lOVic  -  2. 


130.  2 VaT+  2-3V3«-54-V5z  +  1  =  0. 

131.  Vll  -a;+V8-2a;- V21  +  2 a;  =  0. 
Expand: 

132.  (J-x*y-,    (3a*-262)5;    (a* -J  ft1)7* 
(2-$x*)«;    (3 -***)';   (x*  -  x"2)8. 

133.  (a2-i?/8)6;    (2a2+V3a)4;    (V^  +  l  +  a8)6; 
(l  +  2a-£c2-a8)8. 

134.  Expand  to  four  terms 

(l-3a>)~*;    (l-4a;2)~f;    (1  —  fac4)*;    («r- 2*~*>- 

135.  Eind  the  eighty-seventh  term  of  (2x  —  y)90. 

136.  Eesolve  into  partial  fractions 

3-2<c  S-2x  1 


1-3Z  +  2*2'    Q.-x)(l-3xy   1-a;8 
3-2x 


137.   Expand  to  five  terms 


l-3x  +  2x- 


CHAPTER  XXVI. 
LOGARITHMS. 

411.  If  numbers  are  regarded  as  powers  of  ten,  the  expo- 
nents of  the  powers  are  the  Common  or  Briggs  Logarithms  of 
the  numbers. 

If  A  and  B  denote  two  numbers,  a  and  b  their  loga- 
rithms, then  10°  =  A,  106  =  B ;  or,  written  in  logarithmic 
form,  log  A  =  a,  log  B=b. 

412.  The  logarithm  of  a  product  is  found  by  adding  the 
logarithms  of  its  factors. 

For  AX  B  =  10°  X  10&  =  10«+&.         (§  244) 

Therefore,     log  (A  X  B)  =  a  +  b  =  log  A  +  log  B. 

413.  The  logarithm  of  a  quotient  is  found  by  subtract- 
ing the  logarithm  of  the  divisor  from  that  of  the  dividend. 

A      10° 

For  !=S=10°"*-  (§250) 

Therefore,      log  —  =  a  —  b  —  log  A  —  log  B. 
B 

414.  The  logarithm  of  a  power  of  a  number  is  found  by 
multiplying  the  logarithm  of  the  number  by  the  exponent 
of  the  power. 

For  A*  =  (10a)n  =  10™.  (§  251) 

Therefore,      log  An  =  na  =  n  log  A. 


LOGARITHMS.  373 

415.  The  logarithm  of  the  root  of  a  number  is  found  by 
dividing  the  logarithm  of  the  number  by  the  index  of  the 
root. 

For  1/2  =  l/lfr  =  10».  (§  251) 

Therefore,      log^I  =  ^  =  ^A 
n         n 

416.  The  logarithms  of  1,  10,  100,  etc.,  and  of  0.1,  0.01, 
0.001,  etc.,  are  integral  numbers.  The  logarithms  of  all 
other  numbers  are  fractions. 

Since      10°  =      1,  10"1  (=  ^)      =  0.1, 

10*  m    10,  10-2(=TU)    =0.01, 

102  =  100,  io-»(=TXFv*)  =  o.ooi, 

therefore,  log      1  =  0,  log  0.1      =  —  1, 

log    10  =  1,  log  0.01    =-2, 

log  100  =  2,  log  0.001  =  -  3. 

Hence,  the  common  logarithms  of  all  numbers  between 

1  and       10  will  be      0  +  a  fraction, 

10  and     100  will  be      1  -f  a  fraction, 

100  and  1000  will  be      2  +  a  fraction,' 

1  and  0.1      will  be  —  1  +  a  fraction, 

0.1    and  0.01    will  be  —  2  +  a  fraction, 

0.01  and  0.001  will  be  -  3  +  a  fraction. 

417.  If  the  number  is  less  than  1,  the  logarithm  is 
negative  (§  416),  but  is  written  in  such  a  form  that  the 
fractional  part  is  always  positive. 

418.  Every  logarithm,  therefore,  consists  of  two  parts : 
a  positive  or  negative  integral  number,  which  is  called  the 
characteristic,  and  a  positive  decimal  fraction,  which  is 
called  the  mantissa. 


37£  LOGARITHMS. 

Thus,  in  the  logarithm  3.52184,  the  integral  number  3 
is  the  characteristic,  and  the  fraction  .52184,  the  mantissa. 
In  the  logarithm  0.78256  —  2,  which  is  sometimes  written 
2.78256,  the  integral  number  —  2  is  the  characteristic,  and 
the  fraction  .78256  is  the  mantissa. 

419.  If  the  logarithm  has  a  negative  characteristic,  it 
is  customary  to  change  its  form  by  adding  10,  or  a  mul- 
tiple of  10,  to  the  characteristic,  and  then  indicating  the 
subtraction  of  the  same  number  from  the  result. 

Thus,  the  logarithm  2.78256  is  changed  to  8.78256  -  10 
by  adding  10  to  the  characteristic  and  writing  —  10  after 
the  result.  The  logarithm  13.92732  is  changed  to  7.92732 
—  20  by  adding  20  to  the  characteristic  and  writing  —  20 
after  the  result. 

420.  The  following  rules  are  derived  from  §  416. 

Rule  1.  If  the  number  is  greater  than  1,  make  the 
characteristic  of  the  logarithm  one  unit  less  than  the  num- 
ber of  figures  on  the  left  of  the  decimal  point. 

Rule  2.  If  the  number  is  less  than  1,  make  the  char- 
acteristic of  the  logarithm  negative,  and  one  unit  more  than 
the  number  of  zeros  between  the  decimal  point  and  the 
first  significant  figure  of  the  given  number. 

Rule  3.  If  the  characteristic  of  a  given  logarithm  is 
positive,  make  the  number  of  figures  in  the  integral  part  of 
the  corresponding  number  one  more  than  the  number  of 
units  in  the  characteristic. 

Rule  4.  If  the  characteristic  is  negative,  make  the 
number  of  zeros  between  the  decimal  point  and  the  first 
significant  figure  of  the  corresponding  number  one  less  than 
the  number  of  units  in  the  characteristic. 


LOGARITHMS.  875 

Thus,  the  characteristic  of  log  7849.27  is  3 ;  the  char- 
acteristic of  log  0.037  is  -  2,  or  8.00000  -  10.  If  the 
characteristic  is  4,  the  corresponding  number  has  five 
figures  in  its  integral  part.  If  the  characteristic  is  —  3, 
that  is,  7.00000  —  10,  the  corresponding  fraction  has  two 
zeros  between  the  decimal  point  and  the  first  significant 
figure. 

421.  The  mantissa  of  the  common  logarithm  of  any- 
integral  number,  or  decimal  fraction,  depends  only  upon 
the  digits  of  the  number,  and  is  unchanged  so  long  as  the 
sequence  of  the  digits  remains  the  same.  ' 

For,  changing  the  position  of  the  decimal  point  in  a 
number  is  equivalent  to  multiplying  or  dividing  the  num- 
ber by  a  power  of  10.  Its  common  logarithm,  therefore, 
will  be  increased  or  diminished  by  the  exponent  of  that 
power  of  10 ;  and  since  this  exponent  is  integral,  the  man- 
tissa, or  decimal  part  of  the  logarithm,  will  be  unaffected. 

Thus,  271,960  =  10s-48451,  2.7196  =  100-48451, 

27,196  =  104-48451,  0.27196  =  10948451"10, 

2719.6  =  108-48451,  0.027196  =  lO8-484*1-10, 

271.96  =  102-48461,  0.0027196  =  107-48451-10, 

27.196  =  101-48451,  0.00027196  =  106-48461"10. 

One  advantage  of  using  the  number  ten  as  the  base  of 
a  system  of  logarithms  consists  in  the  fact  that  the  man- 
tissa depends  only  on  the  sequence  of  digits,  and  the  char-, 
aeteristic  on  the  position  of  the  decimal  point. 

422.  The  characteristic  of  the  common  logarithm  of  a 
number  can  be  found  by  the  rules  given  in  §  420 ;  but  the 
mantissa  of  the  logarithm  is  found  by  means  of  a  Table  of 
Logarithms. 


376  LOGARITHMS. 

423.  A  Five-place  Table  of  Logarithms.  In  this  table 
(pp.  389-407)  the  vertical  columns  headed  N  contain 
numbers,  the  columns  headed  0,  1,  2,  3,  etc.,  contain  the  loga- 
rithms, and  the  columns  headed  D  contain  the  tabular 
differences  of  the  logarithms.  On  page  389  both  the  char- 
acteristic and  the  mantissa  are  printed.  On  pages  390-407 
the  mantissa  only  is  printed,  and  the  first  two  figures  of 
the  mantissa  are  printed  in  the  left-hand  column  only. 

A  star  prefixed  to  the  last  three  figures  of  a  logarithm 
indicates  that  the  first  two  figures  are  in  the  line  below. 

The  fractional  part  of  a  logarithm  can  be  expressed  only 
approximately,  and  in  a  five-place  table  all  figures  that 
follow  the  fifth  are  rejected.  Whenever  the  sixth  figure  is 
6  or  more,  the  fifth  figure  is  increased  by  1, 

Thus,  if  the  mantissa  of  a  logarithm  written  to  seven  places  is 
6328732,  it  is  written  in  a  five-place  table  53287.  If  the  mantissa  is 
5328751,  it  is  written  in  a  five-place  table  53288. 


To  Find  the  Logarithm  of  a  Given  Number. 

424.  If  the  given  number  consists  of  one  or  two  figures, 
the  logarithm  is  given  on  page  389.  If  zeros  follow  the 
significant  figures,  or  if  the  number  is  a  proper  decimal 
fraction,  the  characteristic  is  determined  by  §  420. 


425.  If  the  given  number  has  three  significant  figures, 
the  number  will  be  found  in  the  column  headed  N  (pp. 
390-407),  and  the  mantissa  of  its  logarithm  in  the  next 
column  to  the  right,  and  on  the  same  line.     Thus, 

Page  390.  log  145  =  2. 16137,  log  14,500  =  4. 16137. 

Page  395.  log  364  =  2.56110,  log  0.0364  =  8.56110  -  10. 

Page  402.  log  716  =  2.85491,  log  0.716    =9.85491-10. 

Page  406.  log  926  =  2.96661,  log  9260     =3.96661. 


LOGARITHMS.  377 

426.  If  the  given  number  has  four  significant  figures,  the 
first  three  figures  will  be  found  in  the  column  headed  N, 
and  the  fourth  figure  at  the  top  of  the  page  in  the  line 
containing  the  figures  0,  1,  2,  3,  etc.  The  mantissa  will  be 
found  in  the  column  headed  by  the  fourth  figure,  and  on 
the  same  line  with  the  first  three  figures.     Thus, 

Page  403.     log     7682  =  3.88547,     log    76.85  =  1.88564. 
Page  406.     log  93,280  =  4.96979,     log  0.9468  =  9.97626  -  10. 

427.  If  the  given  number  has  five  or  more  significant 
figures,  a  process  called  interpolation  is  required. 

Interpolation  is  based  on  the  assumption  that  between 
two  consecutive  mantissas  of  the  table  the  change  in  the 
mantissa  is  directly  proportional  to  the  change  in  the 
number. 

1.  Find  the  logarithm  of  34,237. 

The  required  mantissa  is  (§  421)  the  same  as  the  mantissa  for  3423.7; 
therefore,  it  will  be  found  by  adding  to  the  mantissa  for  3423  seven 
tenths  of  the  difference  between  the  mantissas  for  3423  and  3424. 

The  mantissa  for  3423  is  53441. 

The  difference  between  the  mantissas  for  3423  and  3424  is  12. 

Hence,  the  mantissa  for  3423.7  is  53441  +  (0.7  of  12)  =  53449. 

Therefore,  the  logarithm  of  34,237  is  4.53449. 

2.  Find  the  logarithm  of  0.001576,4. 

The  required  mantissa  is  the  same  as  the  mantissa  for  1576.4. 
The  mantissa  for  1576  is  19756. 

The  difference  between  the  mantissas  for  1576  and  1577  is  27. 
Hence,  the  mantissa  for  1576.4  is  19756  +  (0.4  of  27)  =  19767. 
Therefore,  the  logarithm  of  0.0015764  is  7.19767  —  10. 

3.  Find  the  logarithm  of  32.6708. 

The  required  mantissa  is  the  same  as  the  mantissa  for  3267.08. 
The  mantissa  for  3267  is  51415. 

The  difference  between  the  mantissas  for  3267  and  3268  is  13. 
Hence,  the  mantissa  for  3267.08  is  51415  +  (0.08  of  13)  =  51416. 
Therefore,  the  logarithm  of  32.6708  is  1.51416. 


378  LOGARITHMS. 

428.  When  the  fraction  of  a  unit  in  the  part  to  be  added 
to  the  mantissa  for  four  figures  is  less  than  0.5,  it  is  to  be 
neglected;  when  it  is  0.5  or  more  than  0.5,  it  is  to  be  taken 
as  one  unit. 

Thus,  in  example  1,  §  427,  the  part  to  be  added  to  the  mantissa  for 
3423  is  8.4,  and  the  .4  is  rejected.  In  example  2,  the  part  to  be  added 
to  the  mantissa  for  1576  is  10.8,  and  11  is  added. 

To  Find  the  Antilogarithm ;  that  is,  the  Number 
Corresponding  to  a  Given  Logarithm. 

429.  If  the  given  mantissa  can  be  found  in  the  table, 
the  first  three  figures  of  the  required  number  will  be  found 
in  the  same  line  with  the  mantissa  in  the  column  headed 
N,  and  the  fourth  figure  at  the  top  of  the  column  containing 
the  mantissa. 

The  position  of  the  decimal  point  is  determined  by  the 
characteristic  (§  420). 

1.  Find  the  number  corresponding  to  the  logarithm 
0.92002. 

Page  404.     The  number  for  the  mantissa  92002  is  8318. 
The  characteristic  is  0 ;  hence  (§  420),  the  number  is  8.318. 

2.  Find  the  number  corresponding  to  the  logarithm 
6.09167. 

Page  390.     The  number  for  the  mantissa  09167  is  1235. 

The  characteristic  is  6 ;  hence  (§  420),  the  number  is  1,235,000. 

3.  Find   the   number    corresponding   to   the   logarithm 

2.51055. 

Page  394.     The  number  for  the  mantissa  51055  is  3240. 
The  characteristic  is  2  ;  hence  (§  420),  the  number  is  324. 

4.  Find  the  number  corresponding  to  the  logarithm 
7.50325  -  10. 

Page  394.     The  number  for  the  mantissa  50325  is  3186. 

The  characteristic  is  —  3 ;  hence  (§  420),  the  number  is  0.003186. 


LOGARITHMS.  379 

430.  If  the  given  mantissa  cannot  be  found  in  the  table, 
find  in  the  table  the  two  adjacent  mantissas  between  which 
the  given  mantissa  lies,  and  the  four  figures  corresponding 
to  the  smaller  of  these  two  mantissas  will  be  the  first  four 
significant  figures  of  the  required  number.  If  more  than 
four  figures  are  desired,  they  may  be  found  by  interpolation, 
as  in  the  following  examples  : 

1.  Find  the  number  corresponding  to  the  logarithm 
1.48762. 

Here  the  two  adjacent  mantissas  of  the  table,  between  which 
the  given  mantissa  48762  lies,  are  found  to  be  (p.  394)  48756  and 
48770.  The  corresponding  numbers  are  3073  and  3074.  The 
smaller  of  these,  3073,  contains  the  first  four  significant  figures  of  the 
required  number. 

The  difference  between  the  two  adjacent  mantissas  is  14,  and  the 
difference  between  the  corresponding  numbers  is  1. 

The  difference  between  the  smaller  of  the  two  adjacent  mantissas, 
48756,  and  the  given  mantissa,  48762,  is  6.  Therefore,  the  number 
to  be  annexed  to  3073  is  T\  of  1  =  0.428,  and  the  fifth  significant 
figure  of  the  required  number  is  4. 

Hence,  the  required  number  is  30.734. 

2.  Find  the  number  corresponding  to  the  .logarithm 
7.82326  -  10. 

Here  the  two  adjacent  mantissas  of  the  table,  between  which  the 
given  mantissa  82326  lies,  are  found  to  be  (p.  401)  82321  and  82328. 
The  corresponding  numbers  are  6656  and  6657.  The  smaller  of 
these,  6656,  contains  the  first  four  significant  figures  of  the  required 
number. 

The  difference  between  the  two  adjacent  mantissas  is  7,  and  the 
difference  between  the  corresponding  numbers  is  1. 

The  difference  between  the  smaller  mantissa,  82321,  and  the 
given  mantissa,  82326,  is  5.  Therefore,  the  number  to  be  annexed 
to  6656  is  f  of  1  =  0.7,  and  the  fifth  significant  figure  of  the  required 
number  is  7. 

Hence,  the  required  number  is  0.0066567. 


380  LOGARITHMS. 

Cologarithms. 

431.    The  logarithm  of  the  reciprocal  of   a  number  is 
called  the  cologarithm  of  the  number. 
If  A  denotes  any  number,  then 

colog  A  =  log  j  =  log  1  -  log  A  (§  413)  =  -  log  A  (§  416). 

Hence,  the  cologarithm  of  a  number  is  equal  to  the  loga- 
rithm of  the  number  with  the  minus  sign  prefixed,  which 
sign  affects  the  entire  logarithm. 

In  order  to  avoid  a  negative  mantissa  in  the  cologarithm, 
it  is  customary  to  substitute  for  —  log  A  its  equivalent 
(10  -  log  A)  -  10. 

Hence,  the  cologarithm  of  a  number  is  found  by  sub- 
tracting the  logarithm  of  the  number  from  10,  and  then 
annexing  —  10  to  the  remainder. 

The  best  way  to  perform  the  subtraction  is  to  begin  on 
the  left  and  subtract  from  9  each  significant  figure  of  log 
A  except  the  last,  and  subtract  this  from  10. 

If  log  A  is  greater  in  absolute  value  than  10  and  less 
than  20,  then  in  order  to  avoid  a  negative  mantissa  it  is 
necessary  to  write  —  log  A  in  the  form 
(20  -  log  A)  -  20. 
So  that,  in  this  case,  colog  A  is  found  by  subtracting  log  A 
from  20,  and  then  annexing  —  20  to  the  remainder. 

1.  Find  the  cologarithm  of  4007. 

10.  - 10 

Page  396.  log  4007  =    3.60282 

colog  4007=    6.39718-10 

2.  Find  the  cologarithm  of  103,992,000,000. 

20.  -20 

Page  390.         log  103992000000  =  11.01700 

colog  103992000000  =    8.98300  —  20 


LOGARITHMS. 


381 


If  the  characteristic  of  log  A  is  negative,  then  the  sub- 
trahend, —  10  or  —  20,  will  vanish  in  finding  the  value  of 
colog  A. 

3.   Find  the  cologarithm  of  0.004007. 


Page  396. 


log  0.004007 


10.  - 10 

7.60282  -  10 


colog  0.004007  =    2.39718 

With  practice,  the  cologarithm  of  a  number  can  be  taken 
from  the  table  as  rapidly  as  the  logarithm  itself. 

By  using  cologarithms  the  inconvenience  of  subtracting 
the  logarithm  of  a  divisor  is  avoided.  For  dividing  by  a 
number  is  equivalent  to  multiplying  by  its  reciprocal. 
Hence,  instead  of  subtracting  the  logarithm  of  a  divisor  its 
cologarithm  may  be  added. 


Exercise  136. 
Find  the  logarithm  of : 

1.  6170.         4.    85.76.         7.    0.8694. 

2.  0.617.        5.    296.8.         8.    0.5908. 

3.  2867.         6.    7004.  9.    73,243. 

Find  the  cologarithm  of  ■ 

13.  72,433.  16.   869.278. 

14.  802.376.        17.    154,000. 

15.  15.7643.        18.   70.0426. 

Find  the  antilogarithm  of : 

22.  2.47246.  25.    1.26784. 

23.  7.89081.  26.    3.79029. 

24.  2.91221.  27.    5.18752. 


10.    67.3208. 

11.    18.5283. 

12.    a0042003 

19. 

0.002403. 

20. 

0.000777. 

21. 

0.051828. 

28. 

9.79029  -  10. 

29. 

7.62328  -  10. 

30. 

6.15465  -  10. 

382  LOGARITHMS 

Computation  by  Logarithms. 

1.  Find  the  value  of  x,  if  x  =  72,214  x  0.08203. 

Page  402.  log  72214     =  4.85862 

Page  404.  log  0.08203  =  8.91397  -  10 

By  §412.  logx  =3.77259 

Page  399.  x  =  5923.6 

In  using  a  five-place  table  the  numbers  corresponding  to 
mantissas  may  be  carried  to  five  significant  figures,  and  in 
the  first  part  of  the  table  to  six  figures. 

2.  Find  the  product  of  -  908.4  x  0.05392  X  2.117. 

Page  406.  log  908.4      ==  2.95828" 

Page  398.  log  0.05392  =  8.73175  -  10 

Page  392.  log  2.117      =  0.32572 

Page  390.  2.01575  =  log  103.69.    -  103.69.  Ans. 

When  any  of  the  factors  are  negative,  find  their  logarithms,  with- 
out regard  to  the  signs ;  write  n  after  the  logarithm  that  corresponds 
to  a  negative  number.  If  the  number  of  logarithms  so  marked  is  odd, 
the  product  is  negative ;   if  even,  the  product  is  positive. 

3.  Find  the  value  of  x,  if  x  =  5250  -f-  23,487. 


)8.  log  5250    =  3.72016 

Page  392.  colog  23487  =  5.62917  —  10 

Page  392.  log  x         =  9.34933  -  10  =  log  0.22353. 

.-.  x         =  0.22353 

a     jk  a  a        i        *       •*  7.56  X  4667  X  567 

4.    Find  the  value  of  *,  if  x m  899>1  x  0<00337  X  23435 

Page  403.  log  7.56  =  0.87852 

Page  397.  log  4667  =  3.66904 

Page  399.  log  567  =  2.75358 

Page  405.  colog  899.1  =  7.04619  -  10 

Page  394.  colog  0.00337  =  2.47237 

Page  392.  colog  23435  =  5.63013  -  10 

Page  393.  log  x  =2.44983  =  log  281.73. 

.-.x  =281.73. 


LOGARITHMS.  383 


5.  Find  the  cube  of  376. 

Page  395.  log  376  =  2.57519 

Multiply  by  3  (§  414),  3 

Page  398.  log  3763  =  7.72557  =  log  53159000. 

...  3763  =  53,159,000. 

6.  Find  the  square  of  0.003278. 

Page  394.  log  0.003278    =    7.51561  -  10 

Multiply  by  2  (§  414),  2 

Page  390.  log  0.0032782  =  15.03122  -  20  =  log  0.000010745. 

.-.  0.0032782  =  0.000010745. 

7.  Find  the  square  root  of  8322. 

Page  404.  log  8322  =  3.92023 

Divide  by  2  (§  415),  2  )  3.92023 

Page  406.  log  V8322  =  1.96012  =  log  91.226. 

.-.  V8322  '=91.226. 

If  the  given  number  is  a  proper  fraction,  its  logarithm  will  have  as 
a  subtrahend  10  or  a  multiple  of  10.  In  this  case,  before  dividing  the 
logarithm  by  the  index  of  the  root,  both  the  subtrahend  and  the 
number  preceding  the  mantissa  should  be  increased  by  such  a  number 
as  will  make  the  subtrahend,  when  divided  by  the  index  of  the  root, 
10  or  a  multiple  of  10. 


8.    Find  the  square  root  of  0.000043641. 


Page  396. 

Divide  by  2 
Page  401. 

log  0.000043641 

(§  415), 
log  Vo.  000043641 

=    5.63989- 

10. 
2)15.63989- 
=    7.81995- 

-10 

-  10 
-20 

-  10 

10  =  log  0.0066062. 
...  VO. 000043641  =  0.0066062. 

9.    Find  the  sixth  root  of  0.076553. 

Page  403.         log  0.076553  =    8.88397-10 

50.  —  50 

Divide  by  6  (§  415),  6  )  58.88397  -  60 

Page  400.         log  Vo.076553        =    9.81400  -  10  =  log  0.65163. 
.-.  Vo. 076553        =    0.65163. 


384  LOGARITHMS. 


1A    F    ,,         ■        ,   *  3.1416  X  4771.21  x  2.7183* 
10.   Find  the  value  of  V30,103*  x  0.4343*  x  69.897*' 

log  3.1416    =0.49715  ==  0.49715 

log  4771.21  =  3.67863  =  3.67863 

i log  2.7183    =i  (0.43430)  =  0.21715 

4  colog  30.103    =  4  (8.52139  -  10)  =  4.08556  -  10 

i  colog  0.4343    =£  (0.36221)  =  0.18111 

4  colog  69.897    =  4  (8.15554  -  10)  =  2.62216  -  10 


11.28176  -  20 

30.            -  30 

5)41.28176-50 

8.25635  -  10 

=  log  0.018046. 

11.   Find  the  value  of  x  in  81x  =  10. 

81*  =  10. 

.-.  log  81*  =  log  10. 

x  log  81  =  log  10. 

(§  4H) 

.  =  '<>«  i?  =]-™»  =  0.52397. 

log  81       1.90849 

Exercise  137. 
Find  the  value  of : 

1.  849.7  X  0.7834.  10.  6078  -^-  8703. 

2.  3.709x0.08673.  11.  8.326 -=- 0.1978. 

3.  83.75  X  0.009376.  12.  0.6539  +-  0.9761. 

4.  8593  X  0.0008974.  13.  -  2.567  h-  0.6785. 

5.  -  0.007634  X  6457.  14.  (39.47  X  5.938)  -^  76.54. 

6.  -  0.07843  X  48.66.  15.  (5674  x  0.763)  h-  0.9803. 

7.  -  0.8734  X  0.4378.  16.  357  -f-  (7069  X  0.07948). 

8.  -  7.384  X  (-  5.837).  17.  8.9  h-  (17.81  X  0.002831). 

9.  4657  X  3145.  18.  51.98  -h  (81.71  X  0.0008002). 


LOGARITHMS.  385 

79.32  x  0.005763  x  0.8064 
0.4273  X  0.8462  X  0.01 

72.56  X  0.0005723  X  8979 
42.28  X  4.745  X  0.006158  ' 

0.01723  X  34.^9  X  0.5477 
0.07692  X  37.69  X  0.7733* 

7.126  X  0.7213  X  0.7583 
0.4684  X  7.385  X  0.9673* 

2.057  X  77.12  X  0.004896  X  4.771 
7.582  X  97.33  X  0.008697  X  0.4963* 

24.  5.038.  28.    0.67878.  32.  9.068*. 

25.  15.015.  29.    0.96795.  33.  0.0635*. 

26.  76.854.  30.    0.078678.  34.  0.008721* 

27.  8.7132.  31.    0.0085462,  35.  0.6543*. 

83.25  X  8375  X  0.008576 


19. 
20. 
21. 
22. 
2?. 


3b. 


37. 


38. 


39. 


40. 


0327  X  687.5  X  0.005003 
1632  X  17.74  X  0.7183* 


3.0132  X  34.34  X  0.08137* 
/0.7132  X  9.245  X  0.54772 
*V  76.93  X  0.000173*  X  0.01 

65.02  x  0.002753  X  97.98 
7.2982  X  0.04754  X  8.1562 

3> 


,06012  X  VU6(   2  X  V06( 

\0J 


.5926  x  V0.5926  X  V0.5926 
Find  x  from  the  equation  : 

41.  5*  =  10.  43.    7*  =  40.  45.    (0.4)~*  =  3. 

42.  4*  =  20.  44.    (1.3)*  =  4.2.  46.    (0.9)~*  =  2. 


386  LOGARITHMS. 


Compound  Interest  and  Annuities. 

432.  The  amount  of  $ P  at  compound  interest  at  r  per 
cent 

for  1  year  is  P  (1  +  r) ; 

for  2  years  is  P  (1  +  r)2; 

for  n  years  is  P  (1  -J-  r)n. 

Hence,  if  th'e  amount  for  n  years  is  represented  by  A, 
A  =  P(1  +  r)w. 

Note.     If  the  interest  is  compounded  semi-annually, 
A  =P(1  +  £r)2«. 

Find  the  amount  of  $150  for  6  years  at  4%  compound 
interest 

A  =  P(l  +  r)n  =  150(1.04)6. 
log  150     =2.17609 
log  1.046  =  0.10218 
log  A       =  2.27827  =  log  189.79. 

Hence,  the  required  amount  is  $189.79. 

433.  The  present  worth,  P,  of  $A,  payable  in  n  years  at 
r  per  cent,  must  just  amount  in  n  years  to  A. 

Hence,  F '= ~y,'  (§432) 

434.  An  annuity  is  a  sum  of  money  to  be  paid  at  regular 
intervals  of  time,  as  years,  half  years,  quarter  years. 

435.  To  find  the  present  value  of  an  annuity  of  %A  per  annum 
for  n  years,  at  r  per  cent. 

The  present  value  of  the  1st  payment  is  ;    (§  433) 


LOGARITHMS.  387 


The  present  value  of  the  2d  payment  is    .        r  2 ; 
The  present  value  of  the  rith  payment  is 


(1  4-  r)> 
Hence,  the  present  value  of  all  the  payments  is 

A    +*+& +    - 


.  (1  +  r)       (1  +  r)'  (1  +  rf 

J^\}-\$Tr)\.  (§365) 

1 =— 

1  +  r 
Multiply  both  numerator  and  denominator  by  1  +  r, 


r 

i       ^r(i+r)"-i"i 

"71     (l+r)«     J 


Find  the  present  value  of  an  annuity  of  $500  for  5  years, 
if  money  is  worth  4%. 

ir(l+  r)»  -  1-1  _  500  / 1.045  _xv 
rL    (l  +  r)»    J       0.04  V     1.04*     / 
log  1.045  =  5  X  0.01703  =  0.08515  =  log  1.2166. 

500       0.2166 
*'  0.04  X  1.2166' 

log  500       =  2.69897 

log  0.2166  =  9.33566 -10 
colog  0.04      =  1.39794 
colog  1.2166  =  9.91485  -  10 

3.34742  =  log  2226.5. 

Therefore,  the  present  value  of  the  annuity  is  $2225.50. 


388  LOGARITHMS. 

Exercise  138. 
Find  the  amount  at  compound  interest : 

1.  Of  $8764  for  9  years  at  5%. 

2.  Of  $16,470  for  17  years  at  3£%. 

3.  Of  $12,000  for  12  years  at  4%. 

4.  Of  $976.45  for  9  years  6  months  at  4£%. 
Find  the  principal  that  will : 

5.  Amount  to  $1200  in  7  years  at  5%  compound  interest. 

6.  Amount   to    $18,740  in  12   years   at   4%    compound 

interest. 

7.  Amount  to  $847.55  in  5  years  3  months  at  §\°]0  com- 

pound interest. 

Find  the  rate  of  compound  interest : 

8.  If  $1296  amounts  to  $1576.75  in  5  years. 

9.  If  $4830  amounts  to  $6472.70  in  6  years. 

10.  If  $4625  amounts  to  $7404.80  in  12  years. 

11.  In  what  time  at  3$%  will  $2225  amount  to  $3225  at 

compound  interest  ? 

12.  In  what  time  at  5%  will  $1640  amount  to  $3000, 

interest  being  compounded  semi-annually  ? 

Find  the  present  value  of  an  annuity  : 

13.  Of  $750  for  12  years,  if  money  is  worth  4%. 

14.  Of  $1200  for  10  years,  if  money  is  worth  5J%. 

15.  Of  $1875  for  6  years,  if  money  is  worth  4%. 

16.  Of  $3200  for  14  years,  if  money  is  worth  3£%. 

17.  Of  $2500  for  8  years,  if  money  is  worth  3%. 

18.  Of  $612.50  for  18  years,  if  money  is  worth  3£%. 


A   TABLE 


OOMMON   LOGARITHMS 


NUMBERS 


LOGARITHMS  OF  NUMBERS  TO  100. 


I 

O.OOOOO 

26 

1.41497 

5i 

1.70757 

76 

1. 8808 1 

2 

O.3OIO3 

27 

1 -43 1 36 

52 

1. 71600 

7l 

1.88649 

3 

O.47712 

28 

1.44716 

53 

1.72428 

78 

1.89209 

4 

0.60206 

29 

1.46240 

54 

1.73239 

79 

1.89763 

5 

O.69897 

30 

1.47712 

55 

1.74036 

80 

1.90309 

6 

O.77815 

3i 

1.49136 

56 

1.74819 

81 

1.90849 

7 

O.845 IO 

32 

I-505I5 

57 

I-75587 

82 

1.91381 

8 

O.9O309 

33 

1.51851 

58 

I-76343 

83 

1.91908 

9 

O.95424 

34 

1.53148 

59 

1.77085 

84 

1.92428 

IO 

1. 00000 

35 

1.54407 

60 

I-778I5 

85 

1.92942 

ii 

I.O4I39 

36 

1.55630 

61 

?-78533 

86 

1-9345° 

12 

i. 079 1 8 

37 

1.56820 

62 

1.79239 

!7 

1.93952 

n 

I.  "394 

3* 

1.57978 

63 

J-79934 

88 

1.94448 

14 

1.14613 

39 

1.59106 

64 

1.80618 

89 

1.94939 

15 

1. 17609 

40 

1.60206 

65 

1.81291 

90 

1.95424 

16 

1.20412 

4i 

1. 61278 

66 

1.81954 

9i 

1.95904 

17 

1.23045 

42 

1.62325 

67 

1.82607 

92 

1.96379 

18 

I-25527 

43 

1.63347 

68 

1.83251 

93 

1.96848 

19 

1.27875 

44 

1.64345 

69 

1.83885 

94 

I-973I3 

20 

1.30103 

45 

1.65321 

7° 

1.845 10 

95 

1.97772 

21 

1.32222 

46 

1.66276 

7i 

1.85 126 

96 

1.98227 

22 

1.34242 

47 

1.67210 

72 

1.85733 

97 

1.98677 

23 

I-36I73 

48 

1. 68 1 24 

.  73 

1.86332 

98 

1.99 1 23 

24 

1. 38021 

49 

1 .69020 

74 

1.86923 

99 

1.99564 

25 

1-39794 

50 

1.69897 

75 

1.87506 

100 

2.00000 

390 


100—150 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

IOO 

00000 

043 

087 

130 

173 

217 

260 

303 

346 

389 

43 

IOI 

432 

475 

518 

*6ol 

604 

647 

689 

732 

775 

817 

43 

1 02 

860 

903 

945 

988 

*030 

♦072 

*n5 

*i57 

*i99 

♦242 

42 

103 

01284 

326 

368 

410 

452 

494 

536 

578 

620 

662 

42 

104 

7°3 

745 

787 

828 

870 

912 

953 

995 

*o36 

♦078 

42 

io5 

021 19 

160 

202 

243 

284 

325 

366 

407 

449 

49o 

41 

106 

53i 

572 

612 

653 

694 

735 

776 

816 

857 

898 

4i 

107 

938 

979 

♦019 

*o6o 

*IOO 

*i4i 

*i8i 

*222 

♦262 

*302 

40 

108 

03342 

3^3 

423 

463 

5°3 

543 

583 

623 

663 

703 

40 

109 

743 

782 

822 

862 

902 

941 

981 

*02I 

*o6o 

*IOO 

40 

no 

04139 

179 

218 

258 

297 

336 

376 

415 

454 

493 

39 

III 

532 

57i 

610 

650 

689 

727 

766 

805 

844 

883 

39 

112 

922 

961 

999 

*o38 

*o77 

*ii5 

*i54 

♦192 

♦231 

*269 

39 

"3 

05308 

346 

385 

423 

461 

500 

538 

576 

614 

652 

38 

114 

690 

729 

767 

805 

843 

881 

918 

956 

994 

♦032 

3S   ; 

"5 

06070 

108 

H5 

183 

221 

258 

296 

333 

311 

408 

38 

116 

446 

483 

893 

558 

595 

£>33 

670 

707 

744 

781 

37 

117 

819 

856 

930 

967 

♦004 

♦041 

♦078 

*n5 

♦151 

37 

118 

07188 

225 

262 

298 

335 

372 

408 

445 

482 

5l8 

37 

119 

555 

59i 

628 

664 

700 

737 

773 

809 

846 

882 

36 

120 

918 

954 

99o 

♦027 

♦063 

♦099 

*i35 

*i7i 

♦207 

♦243 

36 

121 

08279 

3H 

35o 

386 

422 

458 

493 

529 

(565 

600 

36 

122 

636 

672 

707 

743 

778 

814 

849 

884 

920 

955 

35 

123 

991 

♦026 

*o6i 

♦096 

♦  132 

♦167 

*202 

*237 

♦272 

*307 

35 

124 

09342 

377 

412 

447 

482 

5i7 

552 

587 

621 

656 

35 

125 

691 

726 

760 

795 

830 

864 

899 

934 

968 

♦003 

35 

126 

10037 

072 

106 

140 

175 

209 

243 

278 

312 

346 

34 

III 

380 

4i5 

449 

483 

5i7 

55i 

585 

619 

653 

687 

34 

721 

755 

789 

823 

857 

890 

924 

958 

992 

♦025 

34 

129 

1 1059 

093 

126 

160 

193 

227 

26l 

294 

327 

361 

34 

I30 

394 

428 

461 

494 

528 

893 

594 

628 

661 

694 

33 

131 

727 

760 

793 

826 

860 

926 

959 

992 

♦024 

33 

132 

12057 

090 

123 

156 

189 

222 

254 

287 

320 

352 

33 

133 

385 

418 

45° 

483 

516 

548 

581 

613 

646 

678 

33 

134 

710 

743 

775 

808 

840 

872 

905 

937 

969 

*OOI 

32 

J35 

13033 

066 

098 

130 

162 

194 

226 

258 

290 

322 

32  ; 

136 

354 

386 

418 

45° 

481 

5i3 

545 

577 

609 

640 

32 

137 

672 

704 

735 

767 

799 

830 

862 

893 

925 

956 

32  i 

138 

988 

♦019 

♦051 

*o82 

*»4 

*i45 

♦176 

♦208 

♦239 

♦270 

3i 

139 

14301 

333 

364 

395 

426 

457 

489 

520 

55i 

582 

3i 

140 

613 

644 

675 

706 

737 

768 

799 

829 

860 

891 

3i 

141 

922 

953 

983 

♦014 

♦045 

♦076 

*io6 

*i3* 

*i68 

*i98 

31  j 

142 

15229 

259 

290 

320 

35i 

381 

412 

442 

473 

503 

31 

H3 

534 

564 

594 

625 

655 

685 

715 

746 

776 

806 

30  ■ 

144 

836 

866 

897 

927 

957 

987 

♦017 

*047 

*o77 

♦107 

3° 

H5 

16137 

167 

197 

227 

256 

286 

3i6 

346 

376 

406 

30  ' 

146 

435 

465 

495 

524 

554 

584 

613 

643 

673 

702 

30 

147 

732 

761 

791 

820 

850 

879 

909 

938 

967 

997 

29 

148 

17026 

056 

085 

114 

143 

173 

202 

231 

260 

289 

29 

149 

319 

348 

377 

406 

435 

464 

493 

522 

551 

580 

29 

N. 

0 

1 

2 

3 

4 

5 

6    7 

8 

9 

D. 

150— 

200 

391 

N. 

0   1  1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

150 

17609 

638 

667 

696 

725 

754 

782 

811 

840 

869 

29 

151 

898 

926 

955 

984 

*oi3 

♦041 

♦070 

♦099 

♦127 

*i56 

29 

I52 

18184 

213 

241 

270 

298 

327 

355 

384 

412 

441 

29 

153 

469 

498 

¥a 

554 

583 

611 

639 

667 

696 

724 

28 

154 

752 

780 

808 

^37 

865 

893 

921 

949 

977 

*oo5 

28 

155 

19033 

061 

089 

117 

145 

173 

201 

229 

257 

285 

28 

156 

312 

340 

368 

396 

424 

45i 

479 

507 

535 

562 

28 

157 

590 

618 

645 

673 

700 

728 

is*> 

783 

838 

28 

158 

866 

893 

921 

948 

976 

*oo3 

♦030 

♦058 

"085 
358 

*II2 

27 

159 

20140 

167 

194 

222 

249 

276 

303 

330 

385 

27 

160 

412 

439 

466 

493 

520 

548 
817 

575 

602 

629 

656 

27 

161 

683 

710 

737 

763 

790 

844 

871 

898 

925 

27 

162 

952 

978 

*oo5 

♦032 

*059 

"085 

*II2 

*I39 

*i65 

*I92 

27 

163 

21219 

245 

272 

299 

325 

352 

378 

405 

43i 

458 

27 

164 

484 

5" 

537 

564 

590 

617 

643 

669 

696 

722 

26 

165 

748 

775 

801 

827 

854 

880 

906 

932 

958 

985 

26 

166 

2201 1 

037 

063 

089 

"5 

141 

167 

194 

220 

246 

26 

167 

272 

298 

324 

35° 

376 

401 

427 

453 

479 

505 

26 

168 

53i 

557 

840 

608 

634 

660 

686 

712 

737 

763 

26 

169 

789 

814 

866 

891 

917 

943 

968 

994 

*OI9 

26 

170 

23045 

070 

096 

121 

147 

172 

198 

223 

249 

274 

25 

171 

3°o 

325 

35° 

376 

401 

426 

452 

477 

502 

528 

25 

172 

805 

578 

603 

629 

654 

679 

7°4 

729 

754 

779 

25 

173 

830 

855 

880 

905 

93o 

955 

980 

♦005 

♦030 

25 

174 

24055 

080 

i°5 

130 

155 

180 

204 

229 

254 

279 

25 

175 

304 

329 

353 

378 

403 

428 

452 

477 

502 

527 

25 

176 

55i 

576 

601 

625 

650 

674 

699 

724 

748 

773 

25 

177 

797 

822 

846 

871 

895 

920 

944 

969 

993 

*oi8 

25 

178 

25042 

066 

091 

"5 

139 

164 

188 

212 

237 

261 

24 

179 

285 

310 

334 

358 

382 

406 

431 

455 

479 

503 

24 

180 

527 

55i 

575 

600 

624 

648 

672 

696 

72b. 

744 

24 

181 

768 

792 

816 

840 

864 

888 

912 

935 

959 

983 

24 

182 

26007 

031 

055 

079 

102 

126 

150 

174 

198 

221 

24 

183 

245 

269 

293 

316 

34o 

364 

387 

411 

435 

458 

24 

184 

482 

5°5 

529 

553 

576 

600 

623 

647 

670 

694 

24 

185 

717 

741 

764 

788 

811 

834 

858 

881 

905 

928 

23 

186 

95i 

975 

998 

*02I 

*045 

*o68 

♦091 

•114 

*i38 

*i6i 

23 

187 

27184 

207 

231 

254 

277 

300 

323 

346 

37o 

393 

23 

188 

416 

439. 

462 

485 

508 

531 

554 

577 
807 

600 

623 

23 

189 

646 

669 

692 

715 

738 

761 

784 

830 

852 

23 

190 

875 

898 

921 

944 

967 

989 

*OI2 

*o35 

♦058 

*o8i 

23 

191 

28103 

126 

149 

I7o 

194 

217 

24O 

262 

285 

307 

23 

192 

330 

353 

375 

398 

421 

443 

466 

488 

5" 

533 

23 

193 

556 

578 

601 

623 

646 

668 

69I 

713 

735 

ns 

22 

194 

780 

803 

825 

847 

870 

892 

914 

937 

959 

981 

22 

195 

29003 

026 

048 

070 

092 

"5 

137 

159 

181 

203 

22 

196 

226 

248 

270 

292 

314 

336 

358 

380 

403 

425 

22 

197 

447 

469 

491 

513 

535 

557 

579 

601 

623 

Si5 

22 

198 

667 

688 

710 

732 

754 

776 

798 

820 

842 

863 

22 

199 

885 

907 

929 

95 x 

973 

994 

*oi6 

♦038 

*o6o 

*o8i 

22 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

392 


200—250 


N. 

0 

1 

2 

3 

4 

5 

6    7 

8 

9 

D. 

200 

30103 

125 

146 

168 

190 

211 

233 

255 

276 

298 

22 

201 

320 

34i 

363 

384 

406 

428 

449 

47i 

492 

5H 

22 

202 

535 

557 

578 

600 

621 

643 

664 

685 

707 

728 

21 

203 

75o 

771 

792 

814 

835 
*o48 

856 

878 

899 

920 

942 

21 

204 

963 

984 

*oo6 

♦027 

♦069 

♦091 

*II2 

*i33 

*i54 

21 

205 

3U75 

197 

218 

239 

260 

281 

302 

323 

345 

366 

21 

206 

387 

408 

429 

45° 

471 

492 

513 

534 

555 

576 

21 

207 

597 
806 

618 

639 

660 

681 

702 

723 

744 

765 

785 

21 

208 

827 

848 

869 

890 

911 

931 

952 

973 

994 

21 

209 

32015 

035 

056 

077 

098 

118 

139 

160 

181 

201 

21 

210 

222 

243 

263 

284 

305 

325 

346 

366 

387 

408 

21 

211 

428 

449 

469 

490 

5Jo 

531 

552 

572 

593 

613 

20 

212 

634 

654 

675 

695 

7i5 

736 

756 

777 

797 

818 

20 

213 

838 

858 

879 

899 

919 

940 

960 

980 

*OOI 

*02I 

20 

214 

33041 

062 

082 

102 

122 

143 

163 

183 

203 

224 

20 

215 

244 

264 

284 

304 

325 

345 

365 

385 

405 

425 

20 

216 

445 

465 

486 

506 

526 

546 

566 

586 

606 

626 

20 

217 

646 

666 

686 

706 

726 

746 

766 

786 

806 

826 

20 

218 

846 

866 

885 

905 

925 

945 

965 

985 

*oo5 

*025 

20 

219 

34044 

064 

084 

104 

124 

H3 

163 

183 

203 

223 

20 

220 

242 

262 

282 

301 

321 

34i 

361 

380 

400 

420 

20 

221 

439 

459 

479 

498 

518 

537 

557 

577 

596 

6l6 

20 

222 

635 

655 

674 

694 

7i3 

733 

753 

772 

792 

8ll 

19 

223 

830 

850 

869 

889 

908 

928 

947 

967 

986 

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19 

224 

35025 

044 

064 

083 

102 

122 

141 

160 

180 

199 

19 

225 

218 

238 

257 

276 

295 

315 

334 

353 

372 

392 

19 

226 

411 

430 

449 

468 

488 

507 

526 

545 

564 

583 

19 

227 

603 

622 

641 

660 

679 

698 

717 

736 

755 

774 

19 

228 

793 

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832 

851 

870 

889 

908 

927 

946 

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19 

229 

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♦003 

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♦040 

♦059 

♦078 

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19 

230 

36173 

192 

211 

229 

248 

267 

286 

305 

324 

342 

19 

231 

36i 

380 

399 

418 

436 

455 

474 

493 

5" 

53o 

19 

232 

549 

568 

586 

605 

624 

642 

661 

680 

698 

717 

19 

233 

736 

754 

773 

791 

810 

829 

847 

866 

884 

903 

19 

234 

922 

940 

959 

977 

996 

♦014 

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♦070 

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18 

235 

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125 

144 

162 

181 

199 

218 

236 

254 

273 

18 

236 

291 

310 

328 

346 

365 

383 

401 

420 

438 

457 

18 

237 

475 

493 

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548 

566 

585 

603 

621 

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18 

238 

658 

676 

694 

712 

73i 

749 

767 

785 

803 

822 

18 

239 

840 

858 

876 

894 

912 

93i 

949 

967 

985 

♦003 

18 

240 

38021 

039 

o57 

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093 

112 

130 

148 

166 

184 

18 

241 

202 

220 

238 

256 

274 

292 

310 

328 

346 

364 

18 

242 

382 

399 

4i7 

435 

453 

47i 

489 

507 

525 

543 

18 

243 

561 

578 

596 

614 

632 

650 

668 

686 

703 

721 

18 

244 

739 

757 

775 

792 

810 

828 

846 

863 

881 

899 

18 

245 

917 

934 

952 

97o 

987 

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♦023 

♦041 

♦058 

♦076 

18 

246 

39094 

in 

129 

146 

164 

182 

199 

217 

235 

252 

18 

247 

270 

287 

305 

322 

340 

358 

375 

393 

410 

428 

18 

248 

445 

463 

480 

498 

515 

533 

550 

568 

585 

602 

18 

249 

620 

637 

655 

672 

690 

707 

724 

742 

759 

777 

17 

... 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

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250—300 


393 


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0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

250 

39794 

811 

829 

846 

863 

881 

898 

915 

933 

950 

17 

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967 

985 

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17 

252 

40140 

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175 

192 

209 

226 

243 

261 

278 

295 

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253 

312 

329 

346 

364 

381 

398 

415 

432 

449 

466 

17 

254 

483 

500 

518 

535 

552 

569 

586 

603 

620 

637 

17 

255 

654 

671 

688 

705 

722 

739 

756 

773 

790 

807 

17 

256 

824 

841 

858 

875 

892 

909 

926 

943 

960 

976 

17 

257 

993 

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♦027 

♦044 

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♦095 

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17 

258 

41 162 

179 

196 

212 

229 

246 

263 

280 

296 

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17 

259 

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347 

363 

380 

397 

414 

430 

447 

464 

481 

17 

260 

497 

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53i 

547 

564 

581 

597 

614 

631 

647 

17 

26l 

664 

681 

697 

714 

73i 

747 

764 

780 

797 

814 

17 

262 

830 

847 

863 

880 

896 

913 

929 

946 

963 

979 

16 

263 

996 

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♦029 

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♦062 

♦078 

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♦127 

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16 

264 

42160 

177 

193 

210 

226 

243 

259 

275 

292 

308 

16 

265 

325 

341 

357 

374 

390 

406 

423 

439 

455 

472 

16 

266 

488 

5°4 

521 

537 

553 

570 

586 

602 

619 

635 

16 

267 

651 

667 

684 

700 

716 

732 

749 

765 

781 

797 

16 

268 

8i3 

830 

846 

862 

878 

894 

911 

927 

943 

959 

16 

269 

975 

991 

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♦024 

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♦056 

♦072 

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♦104 

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16 

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169 

185 

201 

217 

233 

249 

265 

28l 

16 

271 

297 

313 

329 

345 

361 

377 

393 

409 

4o5 

441 

16 

272 

457 

473 

489 

505 

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537 

553 

569 

584 

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16 

273 

616 

632 

648 

664 

680 

696 

712 

727 

743 

759 

16 

274 

775 

791 

807 

823 

838 

854 

870 

886 

902 

917 

16 

275 

933 

949 

965 

981 

996 

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♦028 

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16 

276 

44091 

107 

122 

138 

154 

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185 

201 

217 

232 

16 

277 

248 

264 

279 

295 

311 

326 

342 

358 

373 

389 

16 

278 

404 

420 

436 

45i 

467 

483 

498 

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529 

545 

16 

279 

560 

576 

592 

607 

623 

638 

654 

669 

685 

700 

16 

28o 

716 

73i 

747 

762 

778 

793 

809 

824 

840 

855 

15 

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871 

886 

902 

917 

932 

948 

963 

979 

994 

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15 

282 

45025 

040 

056 

071 

086 

102 

117 

133 

148 

163 

15 

283 

179 

194 

209 

225 
378 

240 

255 

271 

286 

301 

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15 

284 

332 

347 

362 

393 

408 

423 

439 

454 

469 

15 

2ll 

484 

500 

515 

530 

545 

561 

576 

59i 

606 

621 

15 

286 

637 

652 

667 

682 

697 

712 

728 

743 

758 

773 

15 

287 

788 

803 

818 

834 

849 

864 

879 

894 

909 

924 

15 

288 

939 

954 

969 

984 

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15 

289 

46090 

105 

120 

135 

150 

165 

180 

195 

210 

225 

15 

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240 

255 

270 

285 

300 

315 

330 

345 

359 

374 

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389 

404 

419 

434 

449 

464 

479 

494 

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523 

15 

292 

538 

553 

568 

583 

598 

613 

627 

642 

657 

672 

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293 

687 

702 

716 

73i 

746 

761 

776 

790 

805 

820 

15 

294 

835 

850 

864 

879 

894 

909 

923 

938 

953 

967 

15 

295 

982 

997 

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♦026 

♦041 

♦056 

♦070 

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15 

296 

47129 

144 

159 

173 

188 

202 

217 

232 

246 

261 

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297 

276 

290 

305 

319 

334 

349 

363 

378 

392 

407 

15 

298 

422 

436 

451 

465 

480 

494 

5°9 

524 

538 

553 

15 

299 

567 

582 

596 

611 

625 

640 

654 

669 

683 

698 

15 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

6 

9 

D. 

394 


300—350 


N. 

0     1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

300 

47712 
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727 

74i 

756 

770 

784 

799 

813 

828 

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14 

301 

871 

885 

900 

914 

929 

943 

958 

972 

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14 

302 

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015 

029 

044 

058 

073 

087 

101 

116 

130 

14 

303 

144 

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173 

187 

202 

216 

230 

244 

259 

273 

14 

304 

287 

302 

316 

330 

344 

359 

373 

387 

401 

416 

14 

305 

430 

444 

458 

473 

487 

501 

515 

530 

544 

558 

14 

306 

572 

586 

601 

615 

629 

643 

657 

671 

686 

700 

14 

307 

7H 

728 

742 
883 

756 

770 

785 

799 

813 

827 

841 

14 

308 

855 

869 

897 

911 

926 

940 

954 

968 

982 

14 

309 

996 

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♦052 

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14 

310 

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150 

164 

178 

192 

206 

220 

234 

248 

262 

14 

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276 

290 

304 

318 

332 

346 

360 

374 

388 

402 

H 

312 

415 

429 

443 

457 

471 

485 

499 

513 

527 

541 

H 

313 

554 

568 

582 

596 

610 

624 

638 

651 

665 

679 

H 

3H 

693 

707 

721 

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748 

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776 

790 

803 

817 

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3i5 

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886 

900 

914 

927 

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969 

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♦024 

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14 

317 

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120 

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161 

174 

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202 

215 

229 

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3i8 

243 

256 

270 

284 

297 

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338 

352 

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319 

379 

393 

406 

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433 

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461 

474 

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501 

14 

320 

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529 

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610 

623 

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321 

651 

664 

678 

691 

705 

718 

732 

745 

759 

772 

14 

322 

786 

799 

813 

826 

840 

853 

866 

880 

893 

907 

13 

323 

920 

934 

947 

961 

974 

987 

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♦014 

♦028 

♦041 

13 

324 

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068 

081 

095 

108 

121 

135 

148 

162 

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13 

325 

188 

202 

348 

228 

242 

255 

268 

282 

295 

308 

13 

326 

322 

335 

362 

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388 

402 

415 

428 

441 

13 

327 
328 

455 

468 

481 

495 

508 

521 

534 

548 

561 

574 

13 

587 

601 

614 

627 

640 

654 

667 

680 

693 

706 
838 

13 

329 

720 

733 

746 

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772 

786 

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812 

825 

13 

330 

983 

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878 

891 

904 

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930 

943 

957 

97o 

13 

331 

996 

♦009 

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13 

332 

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127 

140 

153 

166 

179 

192 

205 

218 

231 

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333 

244 

257 
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270 

284 

297 

310 

323 

336 

349 

362 

13 

334 

375 

401 

414 

427 

440 

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466 

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13 

335 

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530 

543 

556 

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621 

13 

336 

634 

647 

660 

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686 

699 

711 

724 

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13 

337 

763 

776 

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802 

815 

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853 

866 

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13 

338 

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969 

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♦007 

13 

339 

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046 

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122 

135 

13 

340 

148 

161 

173 

186 

199 

212 

224 

237 

250 

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13 

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275 

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339 

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364 

377 

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13 

342 

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441 

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466 

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517 

13 

343 

529 

542 

555 

567 

593 

605 

618 

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13 

344 

656 

668 

681 

694 

706 

719 

732 

744 

757 

769 

13 

345 

782 

794 

807 

820 

832 

845 

857 

870 

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13 

346 

908 

920 

933 

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13 

347 

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070 

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108 

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145 

13 

348 

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170 

195 

208 

220 

233 

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12 

349 

295 

307 

320 

332 

345 

357 

370 

382 

394 

12 

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0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

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350—400 


395 


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0 

1 

2 

3    4 

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6 

7 

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9 

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419 

432 

444 

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469 

481 

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543 

555 

568 

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617 

630 

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654 

667 

679 

691 

704 

716 

728 
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74i 

753 

765 

12 

353 

777 

790 

802 

814 

827 

839 

864 

876 

888 

12 

354 

900 

913 

925 

937 

949 

962 

974 

986 

998 

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12 

355 

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035 

047 

060 

072 

084 

096 

108 

121 

133 

12 

356 

145 

157 

169 

182 

194 

206 

218 

230 

242 

255 

12 

357 

267 

279 

291 

303 

315 

328 

340 

352 

364 

376 

12 

358 

388 

400 

413 

425 

437 

449 

461 

473 

485 

497 

12 

359 

509 

522 

534 

540 

558 

570 

582 

594 

606 

618 

12 

360 

630 

642 

654 

666 

678 

691 

703 

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727 

739 

12 

361 

75 1 

763 

775 

787 

799 

811 

823 

835 

847 

859 

12 

362 

871 

883 

895 

907 

919 

931 

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955 

967 

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12 

363 

991 

♦003 

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12 

364 

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122 

134 

146 

158 

170 

182 

194 

205 

217 

12 

365 

229 

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253 

265 

277 

289 

301 

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324 

336 

12 

366 

348 

360 

372 

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396 

407 

419 

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12 

367 

467 

478 

490 

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526 

538 

549 

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12 

368 

585 

597 

608 

620 

632 

644 

656 

667 

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12 

369 

703 

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738 

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761 

773 

785 

797 

808 

12 

37° 

820 

832 

844 

855 

867 

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902 

914 

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12 

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937 

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12 

372 

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066 

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101 

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171 

183 

194 

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217 

229 

241 

252 

264 

276 

12 

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287 

299 

310 

322 

334 

345 

357 

368 

380 

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403 

415 

426 

438 

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519 

530 

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600 

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12 

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634 

646 

657 

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115 
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738 

11 

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749 

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11 

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864 

875 

887 

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910 

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967 

11 

380 

978 

990 

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11 

381 

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104 

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127 

138 

149 

161 

172 

184 

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11 

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206 

218 

229 

240 

252 

263 

274 

286 

297 

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11 

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320 

331 

343 

354 

365 

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399 

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11 

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433 

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467 

478 

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512 

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546 

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625 

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11 

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659 

670 

681 

692 

704 

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726 

737 

749 

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11 

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771 

883 

782 

794 

805 

816 

827 

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850 

861 

872 

11 

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906 

917 

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11 

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11 

390 

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118 

129 

140 

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162 

173 

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195 

207 

11 

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218 

229 

240 

251 

262 

273 

284 

295 

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329 

340 

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362 

373 

384 

395 

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472 

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494 

506 

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539 

11 

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550 

561 

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583 

594 

605 

616 

627 

638 

649 

11 

395 

660 

671 

682 

693 

704 

715 

726 

737 

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11 

396 

770 
879 

780 

791 

802 

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824 

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857 

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912 

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141 

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396 


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400 

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217 

228 

239 

249 

260 

271 

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358 

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670 

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713 

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11 

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767 

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810 

821 

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11 

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863 

874 

885 

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917 

927 

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11 

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970 

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119 

130 

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172 

183 

194 

204 

215 

225 

236 

247 

257 

268 

11 

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278 

289 

300 

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342 

352 

363 

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11 

411 

384 

395 

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426 

437 

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479 

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532 

542 

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616 

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658 

669 

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11 

414 

700 

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721 

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742 

752 

763 

773 

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10 

415 

805 

815 

826 

836 

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868 

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10 

416 

909 

920 

930 

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10 

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221 

232 

242 

252 

263 

273 

284 

294 

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420 

428 

335 

346 

356 

366 

377 

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397 

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418 

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439 

449 

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480 

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10 

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542 

552 

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603 

613 

624 

10 

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634 

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655 

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706 

716 

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10 

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737 

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778 

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870 

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900 

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921 

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205 

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246 

256 

266 

276 

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296 

306 

317 

327 

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347 

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397 

407 

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438 

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448 

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498 

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528 

538 

10 

432 

548 

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568 

579 

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10 

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649 

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187 

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256 

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335 

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10 

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464 

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532 

10 

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601 

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443 

640 

650 

660 

670 

680 

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699 

709 

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10 

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738 

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768 

777 

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807 

816 

826 

10 

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836 

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10 

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254 

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273 

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292 

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0 

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450—500 


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45° 

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331 

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10 

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725 

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763 

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801 

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830 

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237 

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398 


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500 

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200 

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252 

260 

269 

278 

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312 

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329 

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398 

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415 

424 

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535 

544 

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603 

612 

621 

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646 

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680 

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714 

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757 

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800 

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817 

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859 

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902 

910 

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105 

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181 

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198 

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240 

248 

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265 

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290 

299 

307 

315 

324 

332 

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349 

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533 

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617 

625 

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684 

692 

700 

709 

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725 

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222 

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255 

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263 

272 

280 

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346 

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362 

370 

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705 

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754 

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770 

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852 

860 

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892 

900 

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207 

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239 

247 

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263 

272 

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296 

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320 

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336 

344 

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360 

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400 

408 

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424 

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464 

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480 

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520 

528 

536 

544 

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8 

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560 

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576 

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600 

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616 

624 

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8 

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640 

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547 

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870 

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550  —  600 


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194 

202 

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233 

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249 

257 

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273 

280 

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296 

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320 

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359 

367 

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702 

710 

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780 

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850 

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313 

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247 

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262 

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290 

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319 

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340 

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369 

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390 

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7 

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519 

526 

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533 

540 

547 

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7 

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611 

618 

625 

633 

640 

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725 

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613 

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218 

225 

232 

7 

620 

239 

246 

253 

260 

267 

274 

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288 

295 

302 

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316 

323 

330 

337 

344 

351 

358 

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372 

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379 

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768 

775 

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358 

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438 

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538 

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578 

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617 

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657 

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657 

757 

763 

770 

776 

783 

790 

796 

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809 

816 

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823 

829 

836 

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875 

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659 

889 

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915 

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105 

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125 

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184 

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197 

204 

210 

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230 

236 

243 

249 

256 

263 

269 

276 

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282 

289 

295 

302 

308 

315 

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328 

334 

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666 

347 

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360 

367 

373 

380 

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393 

400 

406 

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413 

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668 

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549 

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601 

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782 

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6 

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814 

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827 

834 

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847 

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6 

674 

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937 

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097 

104 

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129 

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149 

155 

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168 

174 

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187 

193 

200 

206 

213 

219 

225 

232 

238 

245 

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251 

257 

264 

270 

276 

283 

289 

296 

302 

308 

6 

681 

315 

321 

327 

334 

340 

347 

353 

359 

366 

372 

6 

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378 

385 

39i 

398 

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410 

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423 

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480 

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575 

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601 

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613 

620 

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6 

686 

632 

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658 

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670 

677 

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689 

6 

687 

696 

702 

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721 

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734 

740 

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6 

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759 

765 

77i 

778 

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809 

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6 

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822 

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847 

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866 

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6 

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910 

916 

923 

929 

935 

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691 

948 

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967 

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979 

985 

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998 

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6 

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017 

023 

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036 

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6 

693 

073 

080 

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092 

098 

105 

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117 

123 

130 

6 

994 

136 

142 

148 

155 

161 

167 

173 

180 

186 

192 

6 

695 

198 

205 

211 

217 

223 

230 

236 

242 

248 

255 

6 

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261 

267 

273 

280 

286 

292 

298 

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317 

6 

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323 

330 

336 

342 

348 

354 

361 

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373 

379 

6 

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386 

392 

398 

404 

410 

417 

423 

429 

435 

442 

6 

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448 

454 

460 

466 

473 

479 

485 

49i 

497 

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0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

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402 


700—750 


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0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

700 

84510 

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522 

528 

535 

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6 

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572 

578 

584 

590 

597 

603 

609 

615 

621 

628 

6 

702 

634 

640 

646 

652 

658 

665 

671 

677 

683 

689 

6 

703 

696 

702 

708 

714 

720 

726 

733 

739 

745 

75i 

6 

704 

757 

763 

770 

776 

782 

788 

794 

800 

807 

813 

6 

705 

819 

825 

831 

837 

844 

850 

856 

862 

868 

874 

6 

706 

880 

887 

893 

899 

905 

911 

917 

924 

930 

936 

6 

707 

942 

948 

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960 

967 

973 

979 

985 

991 

997 

6 

708 

85003 

009 

016 

022 

028 

034 

040 

046 

052 

058 

6 

709 

065 

071 

077 

083 

089 

095 

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107 

114 

120 

6 

710 

126 

132 

138 

144 

150 

156 

163 

169 

175 

181 

6 

711 

187 

193 

199 

205 

211 

217 

224 

230 

236 

242 

6 

712 

248 

254 

260 

266 

272 

278 

285 

291 

297 

35! 

303 

6 

713 

309 

315 

321 

327 

333 

339 

345 

352 

364 

6 

714 

37o 

37° 

382 

388 

394 

400 

406 

412 

418 

425 

6 

715 

43i 

437 

443 

449 

455 

461 

467 

473 

479 

485 

6 

716 

491 

497 

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516 

522 

528 

534 

540 

546 

6 

717 

552 

558 

564 

570 

576 

582 

588 

594 

600 

606 

6 

718 

612 

618 

625 

631 

637 

643 

649 

655 

661 

667 

6 

719 

673 

679 

685 

691 

697 

703 

709 

715 

721 

727 

6 

720 

733 

739 

745 

751 

757 

763 
824 

769 

775 

781 

788 

6 

721 

794 

800 

806 

812 

818 

830 

836 

842 

848 

6 

722 

854 

860 

866 

872 

878 

884 

890 

896 

902 

908 

6 

723 

914 

920 

926 

932 

938 

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962 

968 

6 

724 

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6 

725 

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040 

046 

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058 

064 

070 

076 

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088 

6 

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094 

100 

106 

112 

118 

124 

130 

136 

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6 

727 

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159 

165 

171 

177 

183 

189 

195 

20I 

207 

6 

728 

213 

219 

225 

231 

237 

243 

249 

255 

26l 

267 

6 

729 

273 

279 

285 

291 

297 

303 

308 

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320 

326 

6 

730 

332 

338 

344 

350 

356 

362 

368 

374 

380 

386 

6 

731 

392 

398 

404 

410 

4i5 

421 

427 

433 

439 

445 

6 

732 

45i 

457 

463 

469 

475 

481 

487 

493 

499 

5°4 

6 

733 

5io 

516 

522 

5? 

534 

540 

546 

552 

558 

564 

6 

734 

57o 

576 

581 

587 

593 

599 

605 

611 

617 

623 

6 

735 

629 

635 

641 

646 

652 

658 

664 

670 

676 

682 

6 

736 

688 

694 

700 

705 

711 

717 

723 

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735 

741 

6 

737 

747 

753 

759 

764 

77o 

776 
835 

782 

788 

794 

800 

6 

738 

806 

812 

817 

823 

829 

841 

847 

853 

859 

6 

739 

864 

870 

876 

882 

888 

894 

900 

906 

911 

917 

6 

740 

923 

929 

935 

941 

947 

953 

958 

964 

970 

976 

6 

74i 

982 

988 

994 

999 

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♦017 

♦023 

♦029 

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6 

742 

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052 

058 

064 

070 

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081 

087 

093 

6 

743 

099 

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116 

122 

128 

134 

140 

146 

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6 

744 

157 

163 

169 

175 

181 

186 

192 

198 

204 

210 

6 

745 

216 

221 

227 

233 

239 

245 

251 

256 

262 

268 

6 

746 

274 

280 

286 

291 

297 

303 

309 

315 

320 

326 

6 

747 

332 

338 

344 

349 

355 

361 

367 

373 

379 

384 

6 

748 

390 

396 

402 

408 

413 

419 

425 

431 

437 

442 

6 

749 

448 

454 

460 

466 

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477 

483 

489 

495 

500 

6 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

750—800 


403 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

75° 

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512 

518 

523 

529 

535 

54i 

547 

552 

558 

6 

751 

564 

57o 

576 

581 

587 

593 

599 

604 

610 

616 

6 

752 

622 

628 

633 

639 

645 

651 

656 

662 

668 

674 

6 

753 

679 

685 

691 

697 

7°3 

708 

714 

720 

726 

73i 

6 

754 

737 

743 

749 

754 

760 

766 

772 

777 

783 

789 

6 

755 

795 

800 

806 

812 

818 

823 

829 

835 

841 

846 

6 

756 

852 

858 

864 

869 

875 

881 

887 

892 

898 

904 

6 

757 
758 

910 

915 

921 

927 

933 

938 

944 

95° 

955 

961 

6 

967 

973 

978 

984 

990 

996 

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6 

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030 

036 

041 

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058 

064 

070 

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6 

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093 

098 

104 

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116 

121 

127 

133 

6 

761 

138 

144 

150 

156 

161 

167 

173 

178 

184 

190 

6 

762 

195 

201 

207 

213 

218 

224 

230 

235 

241 

247 

6 

763 

252 

258 

264 

270 

275 

281 

287 

292 

298 

304 

6 

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315 

321 

326 

332 

338 

343 

349 

355 

360 

6 

765 

366 

372 

377 

383 

389 

395 

400 

406 

412 

417 

6 

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429 

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440 

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457 

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468 

474 

6 

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480 

485 

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497 

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530 

6 

768 

536 

542 

547 

553 

559 

564 

57o 

576 

581 

587 

6 

769 

593 

598 

604 

610 

615 

621 

627 

632 

638 

643 

6 

770 

649 

655 

660 

666 

672 

677 

683 

689 

694 

700 

6 

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705 

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717 

722 

728 

734 

739 

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6 

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762 
818 

767 

773 

779 

784 

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795 

801 

807 

812 

6 

773 

824 

829 

835 

840 

846 

852 

857 

863 

868 

6 

774 

874 

880 

885 

891 

897 

902 

908 

913 

919 

925 

6 

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930 

936 

941 

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958 

964 

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975 

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6 

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6 

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064 

070 

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092 

6 

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098 

104 

109 

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120 

126 

131 

137 

143 

148 

6 

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154 

159 

165 

170 

176 

182 

187 

193 

198 

204 

6 

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209 

215 

221 

226 

232 

237 

243 

248 

254 

260 

6 

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265 

271 

276 

282 

287 

293 

298 

3°4 

310 

315 

6 

782 

321 

326 

332 

337 

343 

348 

354 

360 

365 

371 

6 

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376 

382 

387 

393 

398 

404 

409 

415 

421 

426 

6 

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432 

437 

443 

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459 

465 

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476 

481 

6 

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487 

492 

498 

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515 

520 

526 

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542 

548 

553 

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575 

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6 

18 

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620 

625 

631 

636 

642 

647 

6 

653 

658 

664 

669 

675 

680 

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697 

702 

6 

789 

708 

713 

719 

724 

730 

735 

741 

746 

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757 

6 

790 

763 
818 

768 

774 

779 

785 

790 

796 

801 

807 

812 

5 

791 

823 

829 

834 

840 

845 

851 

856 

862 

867 

5 

792 

873 

878 

883 

889 

894 

900 

905 

911 

916 

922 

5 

793 

927 

933 

938 

944 

949 

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966 

971 

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5 

794 

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102 

108 

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124 

129 

135 

140 

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146 

151 

157 

162 

168 

173 

179 

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189 

195 

5 

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200 

206 

211 

217 

222 

227 

233 

238 

244 

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255 

260 

266 

271 

276 

282 

287 

293 

298 

304 

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0 

1 

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404 


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0 

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800 

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314 

320 

325 

33i 

336 

342 

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352 

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801 

363 

369 

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380 

385 

390 

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601 

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617 

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5 

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693 

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763 

768 

773 

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784 

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5 

809 

795 

800 

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822 

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838 

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5 

810 

849 

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859 

865 

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5 

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5 

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961 

966 

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5 

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116 

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169 

174 

180 

185 

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222 

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238 

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254 

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265 

270 

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275 

281 

286 

291 

297 

302 

307 

312 

318 

323 

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334 

339 

344 

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365 

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392 

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403 

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519 

524 

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566 

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577 

582 

587 

5 

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598 

603 

609 

614 

619 

624 

630 

635 

640 

5  j 

825 

645 

651 

656 

661 

666 

672 

677 

682 

687 

693 

5 

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698 

703 

709 

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719 

724 

730 

735 

740 

745 

5 

827 

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756 

761 

766 

772 

777 

782 

787 

793 

798 

5 

828 

803 

808 

814 

819 

824 

829 

834 

840 

845 

850 

5 

829 

855 

861 

866 

871 

876 

882 

887 

892 

897 

903 

5 

830 

908 

913 

918 

924 

929 

934 

939 

944 

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955 

5 

831 

960 

965 

971 

976 

981 

986 

991 

997 

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5  1 

832 

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023 

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038 

044 

049 

054 

059 

5 

833 

065 

070 

075 

080 

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091 

096 

101 

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117 

122 

127 

132 

m 

143 

148 

153 

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163 

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169 

174 

179 

184 

189 

195 

200 

205 

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215 

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226 

231 

236 

241 

247 

252 

257 

262 

267 

5  1 

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273 

278 

283 

288 

293 

298 

304 

309 

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3i9 

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324 

330 

335 

340 

345 

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361 

366 

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5 

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381 

387 

392 

397 

402 

407 

412 

418 

423 

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840 

428 

433 

438 

443 

449 

454 

459 

464 

469 

474 

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841 

480 

485 

490 

495 

500 

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516 

521 

526 

5 

842 

53i 

536 

542 

547 

552 

557 

562 

567 

572 

578 

5 

843 

583 

588 

593 

598 

603 

609 

614 

619 

624 

629 

5 

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634 

639 

645 

650 

655 

660 

665 

670 

675 

681 

5 

845 

686 

691 

696 

701 

706 

711 

716 

722 

727 

732 

5 

846 

737 

742 

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752 
804 

758 

763 

768 

773 

778 

783 

5 

847 

788 

793 

799 

809 

814 

819 

824 

829 

834 

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840 

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850 

855 

860 

865 

870 

875 

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886 

5 

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896 

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906 

911 

916 

921 

927 

932 

937 

5 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

850—900 


405 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

850 

92942 

947 

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967 

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125 

131 

136 

141 

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146 

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156 

161 

166 

171 

176 

181 

186 

192 

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197 

202 

207 

212 

217 

222 

227 

232 

237 

242 

5 

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247 

252 

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263 

268 

273 

278 

283 

288 

293 

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298 

303 

308 

313 

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323 

328 

334 

339 

344 

5 

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349 

354 

359 

364 

369 

374 

379 

384 

389 

394 

5 

859 

399 

404 

409 

414 

420 

425 

430 

435 

440 

445 

5 

860 

45° 

455 

460 

465 

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475 

480 

485 

490 

495 

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861 

500 

505 

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515 

520 

526 

53i 

536 

54i 

546 

5 

862 

55i 

556 

561 

566 

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576 

58i 

586 

59i 

596 

5 

863 

601 

606 

611 

6l6 

621 

626 

631 

636 

641 

646 

5 

864 

651 

656 

661 

666 

671 

676 

682 

687 

692 

697 

5 

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702 

707 

712 

717 

722 

727 

732 

737 

742 

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866 

752 

757 

762 

767 

772 

777 

782 

787 

792 

797 

5 

867 

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807 

812 

817 

822 

827 

832 

837 

842 

847 

5 

868 

852 

857 

862 

867 

872 

877 

882 

887 

892 

897 

5 

869 

902 

907 

912 

917 

922 

927 

932 

937 

942 

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870 

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992 

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017 

022 

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062 

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101 

106 

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116 

121 

126 

131 

136 

141 

146 

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874 

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156 

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166 

171 

176 

181 

186 

191 

196 

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201 

206 

211 

216 

221 

226 

231 

236 

240 

245 

5 

876 

250 

255 

260 

265 

270 

275 

280 

285 

290 

295 

5 

877 

300 

305 

310 

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320 

325 

330 

335 

340 

345 

5 

878 

349 

354 

359 

364 

369 

374 

379 

384 

389 

394 

5 

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399 

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409 

414 

419 

424 

429 

433 

438 

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5 

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468 

473 

478 

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498 

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517 

522 

527 

532 

537 

542 

5 

882 

547 

552 

557 

562 

567 

57i 

576 

58i 

586 

59i 

5 

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596 

601 

606 

6ll 

616 

621 

626 

630 

635 

640 

5 

884 

645 

650 

655 

660 

665 

670 

675 

680 

685 

689 

5 

885 

694 

699 

704 

709 

714 

719 

724 

729 

734 

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5 

886 

743 

748 

753 

758 

763 

768 

773 

778 

783 

787 

5 

887 

792 

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802 

807 

812 

817 

822 

827 

832 

836 

5 

888 

841 

846 

851 

856 

861 

866 

871 

876 

880 

885 

5 

889 

890 

895 

900 

905 

910 

915 

919 

924 

929 

934 

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890 

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959 

963 

968 

973 

978 

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998 

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892 

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119 

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134 

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163 

168 

173 

177 

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182 

187 

192 

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202 

207 

211 

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221 

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231 

236 

240 

245 

250 

255 

260 

265 

270 

274 

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897 

279 

284 

289 

294 

299 

303 

308 

313 

318 

323 

5 

898 

328 

332 

337 

342 

347 

352 

357 

361 

366 

371 

5 

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376 

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386 

390 

395 

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405 

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415 

419 

5 

N. 

0 

1 

2 

3 

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5 

6 

7 

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9 

D. 

406 


900—950 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

900 

95424 

429 

434 

439 

444 

448 

453 

458 

463 

468 

5 

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472 

477 

482 

487 

492 

497 

501 

506 

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5 

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521 

525 

530 

535 

540 

545 

550 

554 

559 

564 

5 

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578 

583 

588 

593 

598 

602 

607 

612 

5 

904 

617 

622 

626 

631 

636 

641 

646 

650 

655 

660 

5 

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665 

670 

674 

679 

684 

689 

694 

698 

703 

708 

5 

906 

713 

718 

722 

727 

732 

737 

742 

746 

751 

756 

5 

907 

761 

766 

770 

775 

780 

785 

789 

794 

799 

804 

5 

908 

809 

813 

818 

823 

828 

832 

837 

842 

847 

852 

5 

909 

856 

861 

866 

871 

875 

880 

885 

890 

895 

899 

5 

910 

904 

909 

914 

918 

923 

928 

933 

938 

942 

947 

5 

911 

952 

957 

961 

966 

971 

976 

980 

985 

990 

995 

5 

912 

999 

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♦009 

♦014 

♦019 

♦023 

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♦033 

♦038 

♦042 

5 

913 

96047 

052 

057 

061 

066 

071 

076 

080 

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090 

5 

914 

095 

099 

104 

109 

114 

118 

123 

128 

133 

137 

5 

915 

142 

147 

!52 

156 

161 

166 

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175 

180 

185 

5 

916 

190 

194 

199 

204 

209 

213 

218 

223 

227 

232 

5 

917 
918 

237 

242 

246 

25i 

256 

261 

265 

270 

275 

280 

5 

284 

289 

294 

298 

303 

308 

313 

317 

322 

327 

5 

919 

332 

336 

341 

346 

350 

355 

360 

365 

369 

374 

5 

920 

379 

384 

388 

393 

398 

402 

407 

412 

417 

421 

5 

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426 

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440 

445 

45° 

454 

459 

464 

468 

5 

922 

473 

478 

483 

487 

492 

497 

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506 

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515 

5 

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520 

525 

530 

534 

5£? 

544 

548 

553 

558 

562 

5 

924 

567 

572 

577 

581 

586 

591 

595 

600 

605 

609 

5 

925 

614 

619 

624 

628 

633 

638 

642 

647 

652 

656 

5 

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661 

666 

670 

675 

680 

685 

689 

694 

699 

7°3 

5 

927 

708 

713 

717 

722 

727 

73i 

736 

7il 

745 

75° 

5 

928 

755 

759 

764 

769 

774 

778 

783 

788 
834 

792 

797 

5 

929 

802 

806 

811 

816 

820 

825 

830 

839 

844 

5 

930 

848 

853 

858 

862 

867 

872 

876 

881 

886 

890 

5 

931 

895 

900 

904 

909 

914 

918 

923 

928 

932 

937 

5 

932 

942 

946 

95i 

956 

960 

965 

970 

974 

979 

984 

5 

933 

988 

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♦025 

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5 

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039 

044 

O49 

053 

058 

063 

067 

072 

077 

5 

935 

081 

086 

090 

095 

100 

104 

109 

114 

118 

123 

5 

936 

128 

132 

137 

142 

146 

I51 

155 

l6o 

165 

169 

5 

937 

174 

179 

183 

188 

192 

197 

202 

206 

211 

216 

5 

938 

220 

225 

230 

234 

239 

243 

248 

253 

257 

262 

5 

939 

267 

271 

276 

280 

285 

290 

294 

299 

304 

308 

5 

940 

3i3 

317 

322 

327 

33i 

336 

340 

345 

35o 

354 

5 

941 

359 

364 

368 

373 

377 

382 

387 

39i 

396 

400 

5 

942 

405 

410 

414 

419 

424 

428 

433 

437 

442 

447 

5 

943 

45i 

456 

460 

465 

470 

474 

479 

483 

488 

493 

5 

944 

497 

502 

506 

511 

516 

520 

525 

529 

534 

539 

5 

945 

543 

548 

552 

557 

562 

566 

571 

575 

I8? 

585 

5 

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589 

594 

598 

603 

607 

612 

617 

621 

626 

630 

5 

947 

635 

640 

644 

649 

653 

658 

663 

667 

672 

676 

5 

948 

681 

685 

690 

695 

699 

704 

708 

7i3 

717 

722 

5 

949 

727 

731 

736 

740 

745 

749 

754 

759 

763 

768 

5 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

950—1000 


407 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

95° 

97772 
818 

777 

782 

786 

791 

795 

800 

804 

809 

813 

5 

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823 

827 

832 

836 

841 

845 

850 

855 

859 

5 

952 

864 

868 

873 

877 

882 

886 

891 

896 

900 

905 

5 

953 

909 

914 

918 

923 

928 

932 

937 

941 

946 

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5 

954 

955 

959 

964 

968 

973 

978 

982 

987 

991 

996 

5 

955 

98000 

005 

009 

014 

019 

023 

028 

032 

037 

041 

5 

956 

046  050 

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064 

068 

073 

078 

082 

087 

5 

957 

091 

096 

100 

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109 

114 

118 

123 

127 

132 

5 

958 

137 

141 

146 

150 

155 

159 

164 

168 

173 

177 

5 

959 

182 

186 

191 

!95 

200 

204 

209 

214 

218 

223 

5 

960 

227 

232 

236 

241 

245 

250 

254 

259 

263 

268 

5 

961 

272 

277 

281 

286 

290 

295 

299 

3°4 

308 

3*3 

5 

962 

318 

322 

327 

33* 

336 

340 

345 

349 

354 

358 

5 

963 

363 

367 

372 

376 

38i 

385 

390 

394 

399 

403 

5 

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408 

412 

417 

421 

426 

430 

435 

439 

444 

448 

5 

965 

453 

457 

462 

466 

47i 

475 

480 

484 

489 

493 

4 

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498 

502 

507 

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516 

520 

525 

529 

534 

5lS 

4 

967 

543 

547 

552 

556 

561 

565 

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574 

579 

583 

4 

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588 

592 

597 

601 

605 

610 

614 

619 

623 

628 

4 

969 

632 

637 

641 

646 

650 

655 

659 

664 

668 

673 

4 

970 

677 

682 

686 

691 

695 

700 

704 

709 

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717 

4 

971 

722 

726 

73i 

7P 

740 

744 

749 

753 

758 

762 

4 

972 

767 

771 

776 

780 

784 

789 

793 

798 

802 

807 

4 

973 

811 

816 

820 

825 

829 

834 

838 

843 

847 

851 

4 

974 

856 

860 

865 

869 

874 

878 

883 

887 

892 

896 

4 

975 

900 

905 

909 

914 

918 

923 

927 

932 

936 

941 

4 

976 

945 

949 

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958 

963 

967 

972 

976 

981 

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4 

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989 

994 

998 

♦003 

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4 

978 

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065 

069 

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4 

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092 

096 

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105 

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114 

118 

4 

980 

123 

127 

131 

136 

140 

145 

149 

154 

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162 

4 

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167 

171 

176 

180 

185 

189 

193 

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202 

207 

4 

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211 

216 

220 

224 

229 

233 

238 

242 

247 

25 * 

4 

983 

255 

260 

264 

269 

273 

277 

282 

286 

291 

295 

4 

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300 

304 

308 

3*3 

317 

322 

326 

330 

335 

339 

4 

985 

344 

348 

352 

357 

361 

366 

370 

374 

379 

383 

4 

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388 

392 

396 

401 

405 

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414 

419 

423 

427 

4 

987 

432 

436 

441 

445 

449 

454 

458 

463 

467 

471 

4 

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476 

480 

484 

489 

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498 

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4 

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520 

524 

528 

533 

537 

542 

546 

550 

555 

559 

4 

990 

564 

568 

572 

577 

58i 

585 

590 

594 

599 

603 

4 

991 

607 

612 

616 

621 

625 

629 

634 

638 

642 

647 

4 

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651 

656 

660 

664 

669 

673 

977 

682 

686 

691 

4 

993 

095 

699 

704 

708 

712 

717 

721 

726 

730 

73J 

4 

994 

739 

743 

747 

752 

756 

760 

765 

769 

774 

778 

4 

995 

782 

787 

791 

795 

800 

804 

808 

813 

817 

822 

4 

996 

826 

830 

835 

839 

843 

848 

852 

856 

861 

865 

4 

997 
998 

870 

874 

878 

883 

887 

891 

896 

900 

904 

909 

4 

913 

917 

922 

926 

930 

935 

939 

944 

948 

952 

4 

999 

957 

961 

965 

970 

974 

978 

983 

987 

991 

996 

4 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

D. 

TO  THE  TEACHER 

Many  colleges  now  require  for  entrance  examination  a 
very  elementary  knowledge  of  the  rudiments  of  the  subject 
of  Graphs.  It  is  the  opinion  of  many  good  teachers  that  an 
insight  into  Graphs  is  of  considerable  value  to  the  pupil  in 
finding  the  roots  of  equations,  especially  equations  of  the 
second  degree  and  of  degree  higher  than  the  second.  All 
agree  that  the  study  of  Graphs  tends  to  stimulate  the 
interest  of  the  pupil  in  the  work  of  finding  the  roots  of 
equations. 

At  the  request  of  many  teachers  and  superintendents  this 
short  chapter  has  been  added  to  this  treatise  on  elementary 
algebra.  Five  pages  are  devoted  to  giving  the  necessary 
definitions  and  explanations  of  the  subject,  and  showing 
the  pupil  how  to  plot  points.  Four  pages  then  treat  of 
solving  linear  equations,  six  pages  of  solving  quadratics, 
and  one  page  of  solving  equations  of  degree  higher  than 
the  second. 

A  natural  way  is  to  study  pages  409-417  after  or  with  the 
chapter  on  Simultaneous  Simple  Equations,  and  to  study 
pages  418-423  after  or  with  the  chapters  on  Quadratic 
Equations  and  Simultaneous  Quadratics. 


408 


CHAPTER  XXVII. 
GRAPHS. 

436.  Graphs.  Diagrams,  called  graphs,  are  often  used  to 
show  in  a  concise  manner  variations  in  temperature,  in 
population,  in  prices,  etc.,  etc. 

437.  Variables  and  Constants.  A  number  that,  under  the 
conditions  of  the  problem  into  which  it  enters,  may  take 
different  values  is  called  a  variable. 

A  number  that,  under  the  conditions  of  the  problem  into 
which  it  enters,  has  a  fixed  value  is  called  a  constant. 

Note.  Variables  are  represented  generally  by  the  last  letters  of 
the  alphabet,  x,  y,  z,  etc.  ;  constants,  by  the  Arabic  numerals  and 
by  the  first  letters  of  the  alphabet,  a,  6,  c,  etc. 

438.  Algebraic  Functions.  A  function  of  a  variable  is  an 
expression  that  changes  in  value  when  the  variable  changes 
in  value.  In  general,  any  expression  that  involves  a  vari- 
able is  a  function  of  that  variable.  If  x  is  involved^only  in 
a  finite  number  of  powers  and  roots,  the  expression  is  called 
an  algebraic  function  of  x. 

An  algebraic  function  of  x  is  rational  and  integral  as  regards 
x,  if  x  is  involved  only  in  positive  integral  powers ;  that  is,  in 
powers  and  numerators,  but  not  in  roots  or  denominators. 

Thus,  x2,   ^x'2  +  x,    are  algebraic  functions  of  x;  but  ax, 

3:3  +  4                         1          x 
>Tc  are  not  algebraic  functions  of  x.     Of  — ,  — »  Vx,  2  x  +  a, 

X2    •  X%      X2  +  a2 

j  ax2  +  6x+c,  the  last  three  only  are  rational  integral  functions  of  x. 

a+6 

For  brevity  a  function  of  x  is  represented  by  /(«),  F(x)} 
<l>(x),  each  of  which  is  read  function  x. 

409 


410 


GRAPHS. 


439.  As  an  easy  example  we  may  illustrate  by  a  graph 
the  changes  in  temperature  for  a  day  from  8  a.m.  to  8  p.m. 

The  official  temperatures  for  Boston,  July  17,  1905,  were 
as  follows:  8  a.m.,  71°;  9  a.m.,  72°;  10  a.m.,  73°;  11  a.m., 
77°;  12  m.,  82°;  1  p.m.,  85°;  2  p.m.,  86°;  3  p.m.,  88°;  4  p.m., 
90°;  5  p.m.,  89°;  6  p.m.,  88°;  7  p.m.,  86°;  8  p.m.,  82°. 


^ 

/ 

9 

J 

„0 

3      A 

K 

\ 

ti 

81 

) 

1 

i 

■1 

„ 

1 

A 

o 

n 

° 

} 

■i 

1 

n 

i 

I 

i 

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3 

} 

i 

\ 

t 

( 

H 

Draw  a  horizontal  line  XX'  and  a  line  OF  perpendicular  to  XX'. 
Using  any  convenient  units  of  length,  lay  off  on  XX'  equal  distances 
to  represent  the  hours  and  on  OY  equal  distances  to  represent  degrees 
of  temperature  from  70°  to  90°.  At  each  point  of  division  on  XX' 
draw  a  perpendicular  of  sufficient  length  to  represent  the  temperature 
at  that  hour.  Through  the  upper  ends  of  these  perpendiculars  draw 
the  line  AB.  This  line,  or  graph,  presents  to  the  eye  a  complete  view 
of  the  changes  in  temperature  for  the  day. 


GRAPHS. 


411 


440.  Coordinates.  Let  XX'  be  a  horizontal  straight  line, 
and  let  YY'  be  a  straight  line  perpendicular  to  the  line 
XX'  at  the  point  0.  Any  point  in  the  plane  of  the  lines 
XX'  and  YY'  is  determined  by  its  distance  and  direction 
from  each  of  the  perpendiculars  XX'  and  YY'. 

The  distance  of  a  point  from  YY'  is  measured  from  0  on 
the  line  XX'  and  is  called  the  abscissa  of  the  point.  The 
distance  of  a  point  from  XX'  is  measured  from  0  on  the 
line  YY\  and  is  called  the  ordinate  of  the  point. 

Thus,  the  abscissa  of  Pi  is  OBh  the  ordinate  of  Pi  is  OA i ; 
the  abscissa  of  P2  is  02?2,  the  ordinate  of  P2  is  OA2 ; 
the  abscissa  of  P3  is  OB3,  the  ordinate  of  P3  is  (L4  3 ; 
the  abscissa  of  P4  is  OB±,  the  ordinate  of  P4  is  0.4  4. 


Y 

A, 

r 

1    2 

! 

» 

B, 

0 

?4 

1 

b 

1 

Ri 

1 

1 

1 

* 

1 

H 

P 

1 

/ 

1 

The  abscissa  and  the  ordinate  of  a  point  are  called  the 
coordinates  of  the  point.  The  lines  XX'  and  YY'  are  called 
the  axes  of  coordinates,  or  the  axes  of  reference ;  the  line  XX' 
is  called  the  axis  of  abscissas,  or  the  axis  of  x\  and  the  line 
YY'  is  called  the  axis  of  ordinates,  or  the  axis  of  y.  The 
point  0  is  called  the  origin. 


412  GRAPHS. 

In  general,  an  abscissa  is  represented  by  x,  and  an  ordi- 
nate by  y.  The  coordinates  of  a  point  whose  abscissa  is  x  and 
ordinate  y  are  written  (x,  y).  In  this  notation  the  abscissa 
is  always  written  first  and  the  ordinate  second. 

Thus,  the  point  (4,  7)  is  the  point  whose  abscissa  is  4  and  ordinate  7. 

Abscissas  measured  to  the  right  of  YY'  are  called  positive, 
to  the  left  of  YY'  are  called  negative ;  ordinates  measured 
above  XX'  are  called  positive,  below  XX'  are  called  negative. 

Thus,  in  the  figure  on  page  411  the  point  Pi  is  (8,  5),  the  point  P2 
is  (—6,  3),  the  point  P3  is  (—  4,  —  3),  and  the  point  P4  is  (5,  —  4). 

441.  Quadrants.  The  axes  of  coordinates  divide  the  plane 
of  the  axes  into  four  parts  called  quadrants.  The  quadrant 
XOY  is  called  Quadrant  I,  the  quadrant  X'OY  is  called 
Quadrant  II,  the  quadrant  X'OY  is  called  Quadrant  III, 
and  the  quadrant  XO  Y  is  called  Quadrant  IV. 

Every  point  in  Quadrant  I  has  a  positive  abscissa  and  a 
positive  ordinate  ;  every  point  in  Quadrant  II  has  a  negative 
abscissa  and  a  positive  ordinate ;  every  point  in  Quadrant  III 
has  a  negative  abscissa  and  a  negative  ordinate ;  every  point  in 
Quadrant  IV  has  a  positive  abscissa  and  a  negative  ordinate. 

Hence,  the  signs  of  the  coordinates  of  a  point  show  at  a 
glance  in  what  quadrant  the  point  is  situated. 

442.  Plotting  Points.  It  is  evident  that  if  the  location  of 
a  point  is  known,  the  coordinates  of  that  point  referred  to 
given  axes  may  be  found  easily  by  measurement ;  and  if  the 
coordinates  of  a  point  are  given,  the  point  may  be  readily 
constructed,  or  plotted. 

Thus,  a  convenient  length  is  taken  as  the  unit,  and  the  point  P  is 
found  by  measurement  to  lie  2  units  to  the  right  of  YY'  and  4  units 
below  XX',  and  is,  therefore,  the  point  (2,  —  4). 

Again,  to  plot  the  point  (—5,  2),  a  distance  of  5  units  is  laid  off  on 
XX'  to  the  left  from  0  to  JTi,  and  a  distance  of  2  units  on  YY'  upwards 
from  O  to  K2.  The  intersection  of  the  perpendiculars  erected  at  K\  and 
K2  determines  the  point  K,  which  is  the  required  point  (—  5,  2). 


GRAPHS. 


413 


Coordinate  paper  is  paper  ruled  in  small  squares.  In 
plotting  points  and  graphs  the  student  will  find  coordinate 
paper  of  much  help  in  giving  accuracy  and  in  saving  time. 


\j 

1 

1 

F 

< 

M 

\ 

( 

S 

K 

V 

h 

J 

c 

91 

F 

X' 

bl 

K 

F 

L 

P 

> 

' 

Exercise  139. 

1.  In  the  figure  on  this  page  determine  the  coordinates 
of  the  point  B ;  of  M ;  of  N\  of  R  ;  of  S)  of  If;  of  L  ;  of 
A  j  of  >j  of  D;  of  (7. 

2.  What  is  the  abscissa  of  a  point  on  the  axis  of  y  ? 
What  is  the  ordinate  of  a  point  on  the  axis  of  x  ? 

3.  Where  must  a  point  lie  if  its  ordinate  is  zero  ?  if  its 
abscissa  is  zero  ?   if  both  abscissa  and  ordinate  are  zero  ? 

4.  Plot  the  following  points  :  (2,  5),  (-  3,  6),  (-  2,  -  4), 
(3,  -  5),  (7,  0),  (-  5,  0),  (0,  0),  (0,  -  3),  (-  4,  -  5),  (7,  2). 

5.  In  what  quadrant  does  a  point  lie  if  its  coordinates  are 
both  positive  ?  if  both  are  negative  ?  if  the  ordinate  is  pos- 
itive and  the  abscissa  negative  ?  if  the  abscissa  is  positive 
and  the  ordinate  negative  ? 


414  GRAPHS. 

6.  Plot  the  points  (-2,  -8),  (-1,  -6),  (0,  -4),  (1,  -2), 
(2,  0),  (3,  2),  (4,  4).  Do  these  points  lie  in  a  straight  line  ? 
Is  the  equation  2  cc  —  ?/  =  4  satisfied  if  the  abscissas  are  sub- 
stituted in  turn  for  x,  and  the  corresponding  ordinates  for  y  ? 

443.  Graph  of  a  Function.  Let  f(x)  be  an  algebraic  function 
of  x,  where  x  is  a  variable.  If  y  =f(x),  then  ?/  is  a  new  vari- 
able connected  with  x  by  the  relation  of  y  =f(x).  If  /(#)  is 
rational  and  integral,  it  is  evident  that  to  every  value  of  x 
there  corresponds  one  value,  and  only  one  value,  of  y. 

If  different  values  of  x  are  laid  off  as  abscissas,  and  the 
corresponding  values  of  f(x)  as  ordinates,  a  series  of  points 
will  be  obtained.  A  line,  straight  or  curved,  may  be  drawn 
through  all  these  points.  This  line  is  called  the  graph  of  the 
function  f(x) ;  it  is  also  called  the  graph  of  the  equation  y  =  f(x) . 

Plot  the  graph  of  the  equation  x  —  2  y  —  4  =  0. 
Transpose,  2  y  =  x  —  4. 

05-4 


The  following  table  may  be  computed  readily. 


If  x  =  0,  y  =  -2 
x=-2,  y  =  -S 
s  =  -4,        ?/  =  -  4 


If  x  =  +  2,        y  =  -  1 ; 

x=+4,         y=      0; 
x  =  +  6,        y  =  +  1  j 


x  =  -6,        y  =  -5.  x  =  +  8,        y  =  +2. 

These  points  are  plotted  in  the  figure  on  page  415  and  all  lie  on  the 
►straight  line  A B.  If,  in  the  given  equation  x— 2y  —  4=0,  the  abscissa 
of  any  point  in  the  line  AB  is  substituted  for  x  in  the  corresponding 
ordinate  for  y,  the  equation  is  satisfied.  The  line  AB  extends  indefi- 
nitely in  either  direction  and  is  the  graph  of  the  equation  x— 2  y— 4=0. 

If  any  two  points  of  a  straight  line  are  known,  the  position 
of  the  line  is  definitely  determined.    ■ 

444.  Linear  Equations.  The  graph  of  every  equation  of 
the  form  ax  +  by  +  c  =  0  is  a  straight  line.  For  this  reason 
such  an  equation  is  often  called  a  linear  equation. 


GRAPHS. 


415 


> 

0 

1 

Exercise  140. 

Plot  the  graph  of  the  following  equations  by  finding  a 
series  of  points: 

1.  Sx-2y  =  6.  4.    -x  +  3y  =  6. 

2.  5x  +  2y  =  10.  5.   3x  +  2y  =  12. 

3.  4#  —  y  +  4  =  0.  6.    #  —  5  ?/  =  5. 

Plot  the  graphs  of  the  following  equations  by  finding 
the  points  in  which  the  graphs  cut  the  axes : 

7.  lx  +  2y-lA  =  0.  10.    4x  +  3y  +  12  =  0. 

8.  5x- 32/- 15  =  0.  11.    a; -8?/  + 8  =  0. 

9.  3a;  -4,7/  -24  =  0.  12.    5x  +  ±y  +  30  =  0. 

Plot  the  graphs  of  the  following  equations  by  finding  any 
two  points : 

13.  x  +  y  =  0.        15.    x  —  5y  =  0.       17.    5x  +  4:y  =  Q. 

14.  x  —  y  =  0.        16.    2  x  =  6  2/.  18.    7  x  —  5  y  =  0. 


416 


GRAPHS. 


19.  In  what  respect  do  the  equations  of  Examples  1-12 
differ  from  the  equations  of  Examples  13-18  ? 

20.  Does  the  graph   of  the  equation  ax  ±  by  —  0  pass 
through  the  origin  ?     Why  ? 

21.  The  equation  of  the  axis  XX'  is  y  —  0.     What  is 
the  equation  of  the  axis  YY'? 

22.  What  is  the  position  of  a  graph  if  its  equation  does 
not  contain  x  ?    if  its  equation  does  not  contain  y  ? 

Plot  the  graph  of : 

23.  3x  =  6.  25.    x=-\.  27.    5#  =  30. 

24.  2  y  =  5.  26.    y  =  —  f .  28.    6  y  =  —  42. 


r 

^                                                                         f^ 

^*v*»»^l^                                                          y^ 

^*s*"»««^                                        J 

■^^^                    >' 

"**■*■•      ^K 

/^-^.^ 

^_ o ^ --^ *■ 

/ 

A 

/ 

/ 

f 

/ 

-i/V 

1r    t  r- 

" 

445.  Graph  of  the  Solution  of  a  Pair  of  Simultaneous  Linear 
Equations.  In  the  figure  the  straight  line  AB  is  the  graph 
of  the  equation  x  -f-  3  y  —  12;  and  the  straight  line  CD  is  the 
gr#ph  of  the  equation  4  x  —  3  y  =  18.  It  is  evident  that 
the  coordinates  of  K,  the  point  of  intersection  of  the  lines 
AB  and  CD,  must  satisfy  both  equations. 


GRAPHS.  417 

By  solving  the  equations  as  simultaneous  equations  we 
find  that  x  =  6  and  y  =  2,  which  are  the  coordinates  of  K. 

Hence,  it  is  evident  that  it  is  possible  by  the  use  of  graphs 
to  solve  two  simultaneous  linear  equations  that  contain  only 
two  unknown  numbers.  In  some  cases  exact  values  of  the 
unknown  numbers  may  be  found;  in  other  cases  only  ap- 
proximate values.  The  larger  the  scale  used  in  plotting  the 
graphs,  the  closer  will  be  the  approximations  obtained. 

Exercise  141. 

Find  by  graphs  exact  values  of  x  and  y  in  the  following 
equations  and  verify  by  solving  the  equations  : 

1.  2x-5y  =    0\  4.    llx-2y  =      21\ 
4x  +  2y  =  24:J  2x  +  4:y=-18J 

2.  7x-2y  =  U\  5.    5x  +  $y  =      20\ 
5x+    y  =  10J  2x-3y  =  -23) 

3.  5x  +  ±y  =      301  6.    3x-f  4y  =  30\ 

x-     y  =  -    3J  (   5x-6y  =  12J 

Find  by  graphs  approximate  values  of  x  and  y : 

7.  4:x-5y  =  10^  9.    7x-2y*=U\ 
2x  +  3y=    9J  5x  +  32/  =  15J 

8.  8x+     y=2(Tl  10.    9x-4y  =  l$\ 
2x-5y  =  10)  2x  +  5y  =  20J 

11.  The  graphs  of  the  equations  2x  +  3y=4:,2x— y=12, 
and  x  -f  3  y  =  —  1  meet  in  a  point.  Are  the  equations  simul- 
taneous ?    Give  reason. 

12.  Do  the  graphs  of  the  equations  Ax  —  y  =  2,  x  —  6y  =  5, 
and  3x  +  y  =  10  meet  in  a  point  ?  Are  the  equations  simul- 
taneous ?    Are  the  equations  inconsistent  ? 

13.  Are  the  equations  2x  —  3y  =  5  and  2x  —  3y  =  S 
simultaneous  ?    What  is  shown  by  their  graphs  ? 


418 


GRAPHS. 


446.  Graphs  of  Quadratic  Equations.  The  graph  of  any  given 
quadratic  equation  in  x  and  y  may  be  drawn  by  the  use  of 
the  method  shown  in  the  solution  of  the  following  example. 

Plot  the  graph  of  the  function  x2  +-  3  x  —  4. 

If  y  =  0,  x  =  +  1  or  -  4  ; 
?/  =  +  l,  x  = +  1.19  or  -4.19; 
y=  +  2,  x  =  +  1.37  or  -4.37; 
y  =  +  3,  x  =  +  1.54  or  -  4.54  ; 
y  =  +  4,  x  =  +  1.70  or  -  4.70  ; 
y  =  +  6,  x  =  +  2  or  -  5  ; 
y  =  +  8,  x  =  +  2.27  or  -  5.27. 


it  x2  +  3  a 

4  =  y. 

Then  x  = 

y  = 

-<H, 

X 

=  -1.5; 

y  = 

-6, 

X 

=  -1 

or 

-2; 

y  = 

-5, 

X 

=  -0.38 

or 

-2.62 

y  = 

-4, 

X 

=      0 

or 

-3; 

y  = 

—  3, 

X 

=  +  0.30 

or 

-3.30 

y  = 

-2, 

X 

=  +  0.56 

or 

-3.56; 

y  = 

-1, 

X 

=  +  0.79 

or 

-3.79. 

A 

V 

B 

J 

/ 

I 

/ 

/ 

/ 

1 

J 

C 

* 

o 

D 

/ 

\ 

/ 

\ 

/ 

\ 

V 

I 

\ 

-' 

/ 

^ 

Plot  the  points  found  (-1.5,  -6J),  (-1,  -6),  (-2,  -6),  and  so  on. 
Through  these  points  with  a  free  hand  draw  the  smooth  curve  AB.  The 
curve  AB  is  the  graph  of  the  function  x2  +  3  x  -  4.  This  graph  consists 
of  one  symmetrical  branch  of  infinite  length  and  is  called  a  parabola.  For 
values  of  y  less  than  —  6£,  the  corresponding  values  of  x  are  imaginary. 

When  y  =  0,  then  x  =  1  and  -  4,  the  roots  of  the  equation 
x2  -  3  x  -  4  =  0. 


GRAPHS.  419 

To  solve  an  equation  in  x  it  is  necessary  only  to  find  the 
points  in  which  the  graph  cuts  the  axis  of  x.  The  abscissas 
of  these  points  are  the  roots  of  the  given  equation. 

447.  A  more  rapid  method  of  solving  a  quadratic  "by  the  use 
of  graphs  is  shown  in  the  solution  of  the  following  equation. 
Solve  the  equation  x2  —  x  —  2  =  0. 

Let  x2=y,  and  put  y  for  x2.    Then  the  equation  becomes  y — x — 2 = 0. 

The  graph  of  the  equation  y  —  x  —  2  =  0  is  the  straight  line  AB,  as 
shown  in  the  left-hand  figure  on  page  420,  and  the  graph  of  the  equa- 
tion x2  =  y  is  the  parabola  CD. 

The  abscissas  of  the  intersections  of  AB  and  CD  are  2  and  —  1,  which 
are  the  roots  of  the  given  equation,  as  may  be  shown  by  solving  it. 

The  great  advantage  of  this  method  is  that  the  same 
parabola  may  be  used  in  the  solution  of  different  equations 

Exercise  142. 
Plot  the  graphs  of  the  following  functions : 

1.  x2  +  5  x  +  4.      3.  x2  -  7  x  -f  6.        5.  2  x2  -  7  x  +  5 

2.  x2  +  x-2.         4.  x2-5x-\-6.        6.  3«24-4a;--4 

Find,  by  the  method  of  graphs,  the  roots  of : 

7.  x*  _  x  _  6  =  0.  10.  7  x2  +  14  x  -  21  =  0. 

8.  2  x2  -  9  x  +  9  =  0.  11.  4  x2  -  12  x  +  9  =  0. 

9.  5z2-20  =  0.  12.  25x2  +  60x  +  36  =  0 

Find  approximately  the  roots  of : 

13.  3  x2  -  2  x  -  7  =  0.  15.  5  a2  -  7  a;  -  1  =  0. 

14.  5  ar2  -  3  x  -  30  =  0.  16.  7  a2  +  5  x  -  31  =  0. 

17.  What  is  the  nature  of  the  roots  of  the  equation 
3rc2-r-2:E  +  4  =  0?    Construct  the  graph  of  the  equation. 

18.  How  does  the  graph  of  a  quadratic  show  that  the 
roots  are  real  and  unequal  ?  real  and  equal  ?  imaginary  ? 


420 


GRAPES. 


448.    Solution   of   Simultaneous   Quadratic   Equations.     In 

general,  the  method  of  solving  simultaneous  linear  equa- 
tions (p.  416,  §  445)  should  be  followed. 


Solve 

*2+     2/2  =  251 
4  x    -f  3  y   =  20  J 

(1) 
(2) 

In  (1),  if  x  =      0, 

x  =  ±  1, 
x=±2, 

y  =  ±5;                   x  =  ±3, 
y  =  ±4.90;             x  =  ±  4, 

y  =±4.58;             x  =  ±  5, 

2/  =  ±4; 
y  =  ±3; 

y=    o. 

If  x  >  +  5  or  <  —  5,  the  value  of  y  is  imaginary. 

Equation  (1)  is  symmetrical ;  its  graph  also  is  symmetrical  and  is 
the  circle  HK,  as  shown  in  the  right-hand  figure.  The  graph  of  equa- 
tion (2)  is  found  to  be  the  straight  line  AB  intersecting  the  circle  at 
the  points  -M"(lf,  4|)  and  N(5,  0). 

Hence,  the  solution  gives  x  =  5,  y  =  0orx=  lg,  y  =  4f. 

The  straight  line  CD  is  the  graph  of  4 x  -f  3y  =  25.  (3) 

The  solution  of  (1)  and  (3)  gives  the  double  solution  x  =  4,  y  —  3. 

The  straight  line  EF  is  the  graph  of  4  x  +  3  y  =  30.  (4) 

The  solution  of  (1)  and  (4)  gives  imaginary  roots,  since  the  graphs 
do  not  intersect. 


GRAPHS. 


421 


Exercise  143. 

1.  What  does  the  right-hand  figure  on  page  420  show  about 
the  relation  of  AB,  CD,  and  EF  ?  How  do  the  coefficients 
of  x  and  y  in  equations  (2),  (3),  and  (4)  show  this  ?  Are  the 
graphs  of  ax  -f  by  -f  c  =  0  and  ax  -f-  by  -f  d  =  0  parallel  ?    . 

2.  Write  the  equations  of  two  parallel  lines  and  con- 
struct their  graphs. 


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3.  If  a  straight  line  and  a  circle  touch  each  other,  how  many 
values  has  x  ?  how  many  has  y  ?  How  many  values  have  x 
and  y  when  the  line  cuts  the  circle  ?  What  is  the  nature  of 
the  roots  of  two  equations  when  the  graphs  do  not  intersect  ? 

Solve  exactly  or  approximately  by  the  method  of  graphs  : 


4.       x2  +     if  -  1G9  = 
3x  ~2y  +      9 

6.       x2  +  y2  =  10(n 
5x  -f  y  =    46  J 


2} 


a-2+     y- 


100\ 


3  a;   +4v/  ■  :    50  J 

7.       a-2  4-     y2  =  1001 
3x  +  4y  =    60  J 


422 


GRAPHS. 


8.  Solve  the  equations  of  Example  7  as  simultaneous 
equations  and  explain  why  their  graphs  do  not  intersect. 

9.  The  figure  on  page  421  shows  the  graph  of  the  ellipse 
4  x2  +  9  y2  =  288,  and  the  graph  of  the  parabola  3  y2  =  8  x. 
What  roots  satisfy  these  equations  ? 

10.  The  equation  of  the  circle  ax2  +  ay2  =  c  differs  in  what 
respect  from  the  equation  of  the  ellipse  ax2 +  by2  =  c?  What 
is  the  shape  of  the  ellipse  when  a  and  b  differ  greatly  in  value  ? 
when  a  and  b  are  nearly  equal  ?   when  a  and  b  are  equal  ? 


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11.  The  figure  on  this  page  shows  the  graph  of  the  circle 
x2  +  y2  =  26,  and  the  graph  of  the  hyperbola  #y  =  5.  What 
are  the  coordinates  of  their  points  of  intersection  ?  What 
roots  satisfy  the  equations  ? 

12.  Solve  the  equations  x2  -f  y2  =  26  and  xy  =  5  as  simul- 
taneous quadratics  and  notice  that  the  results  are  the  answers 
to  Example  11. 


GRAPHS. 


423 


Solve  by  graphs : 

13.  x2  +  y1  =  80 

xy  =  32 

14.  x2  +  i 

xy 

15. 
3 


} 

2/2=34| 

y  =15J 

*2  +  2/2  =    74-^ 

z2  +  2/2  =  172j 


16. 


17. 


■>y2 


5x2  +  y2  =  S21\ 
-196  a  =      0J 


^2  +  62/2  =  791 
a2  -  4  y2  =  89  J 

:1) 


18.   Sx2-5y2 
xy 


19.  Explain  the  meaning  of  the  imaginary  roots  of  the 
equations  in  Example  18. 

Note.  The  equations  in  Example  18  are  types  of  the  two  simple 
hyperbola  equations.  The  difference  in  form  is  due  to  the  difference 
in  the  position  of  the  curves. 

Determine  by  inspection  the  shape  of  the  graph  of : 

20.  2x  —  lly=7.     23.    x2  +  y2  =  18.     26.    4z2+4y2=27 

21.  5x2+8y2  =  6.     24.    x  =  3y.  27.    xy  =  12. 

22.  y2  =  8x.  25.    2x2  =  7y.       28.    2x2-5y2=12 


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424 


GRAPHS. 


449.  Graphs  of  Higher  Equations.  The  graphs  of  equations 
and  functions  of  higher  degree  than  the  second  may  be  plotted 
by  the  method  already  shown  (p.  418,  §  446). 

In  general,  the  number  of  real  roots  of  an  equation  in  x 
is  equal  to  the  number  of  times  the  graph  cuts  the  axis  of  x. 
If  the  graph  is  tangent  to  the  axis  of  x,  there  is  a  double 
root  or  a  multiple  root ;  if  the  graph  does  not  cut  or  touch 
the  axis  of  x,  the  roots  are  imaginary. 


Plot  the  graph  of  the  function  xz 
Put  x3  -  x2  -  16  x  -  20  =  y. 


16 


20. 


x  =  +  6, 

y  =  +  64 ; 

If  x  =  +  0.5, 

y  =  -28.13 

x  =  +  5.5, 

y  =  +  28.13; 

x  =      0, 

V  =-20; 

x  =  +  5, 

y=        0; 

x  =  -0.5, 

?/  =  -  12.38 

x  =  +4.5, 

2/  =  -21.13; 

x  =  -l, 

2/  =  -    6; 

x  =+4, 

V  =  -  86 ; 

x=-1.5, 

2/  =  -    1.63 

x  =  +  3.5, 

y  =  -  45.38 ; 

x  =  -  2, 

2/=        0; 

x  =  +3, 

y  =  -  50  ; 

x  =  -2.5, 

2/  =  -    1.88 

x  =  +2.7, 

y  =;  -60.81; 

x  =  -3, 

?  =  -    8; 

x  =  +  2.5, 

y  =  -  50.63  ; 

x  =  -3.5, 

y  =  -19.13 

x  =  +2, 

2/=-48; 

x  =  -4, 

y  =-36; 

x=  +  1.5, 

2/ =-42.88; 

x  =  -4.5, 

2/  =  -  59.38 

x  =  +l, 

y  =  -  36. 

x  =  —  5, 

y  =  -  90. 

To  make  the  figure  compact  use  two  spaces  of  the  coordinate  paper 
for  one  unit  of  x,  and  one  space  for  ten  units  of  y.  The  curve  CAEBD 
(p.  423)  is  the  graph  of  the  function  x3 — x2  — 16  x — 20.  The  graph  shows 
that  the  roots  of  the  equation  x3— x2  — 16  x— 20=0  are  5,  —2,  and  —2. 

When  x  is  greater  than  6  the  curve  evidently  extends  indefinitely 
above  XX' ;  when  x  is  less  than  —  5,  indefinitely  below  XX'. 

To  determine  more  accurately  the  shape  of  the  curve,  it  is  often 
desirable  to  assume  for  x  several  values  between  two  consecutive  units. 


Exercise  144. 

Find  by  a  graph  the  roots  of ; 
1.  ^3  _  x2  __  12  x  =  0.  2.  2  xs  -  x2  -  26  a.  +  40  =  0. 

Find  by  a  graph  the  number  of  real  roots  of  • 
3.  x9  -  8  =  0.  4.  xs  -  5  x2  +  8  a  +  14  =  0. 


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